Mark the decimal point two places from right to left
Mark the decimal point one place from right to left
9/10 = 0.9
4/100
Mark the decimal point two places from right to left
4/100
= 0.04
Convert 2/10 and 6/1000 into decimals
2/10
Mark the decimal point one place from right to left
2/10
= 0.2
6/1000
Mark the decimal point three places from right to left
6/1000
= 0.006
23, 0.2, 0.006 are unlike decimals. So we convert them into like decimals.
Question: 2
Convert each of the following into fractions in the lowest form:
(i) 0.04
(ii) 2.34
(iii) 0.342
(iv) 17.38
Solution:
(i) 0.04
= 0.04/1
= 4/100
= 1/25
(ii) 2.34
= 2.34/1
= 234/100
= 117/50
(iii) 0.342
= 0.342/1
= 342/1000
= 171/500
(iv) 17.38
=17.38/1
= 1738/100
= 869/50
Question: 3
Express the following fractions as decimals:
(i) 23/10
Solution:
(i) 23/10
= 23/10
= 2.3
Question: 4
Add the following:
(i) 41.8, 39.24, 5.01 and 62.6
(ii) 18.03, 146.3, 0.829 and 5.324
Solution:
(i) 148.65
(ii) 170.483
Question: 5
Find the value of:
(i) 9.756 – 6.28
(ii) 48.1- 0.37
(iii) 108.032 – 86.8
(iv) 100 – 26.32
Solution:
(i) 3.476
(ii) 47.73
(iii) 21.232
(iv) 73.68
Question: 6
Take out 3.547 from 7.2
Solution:
3.653
Question: 7
What is to be added to 36.85 to get 59.41?
Solution:
x + 36.85 = 59.41
x = 59.41-36.85
x = 22.56
Therefore 22.56 is added to 36.85 to get 59.41
Question: 8
What is to be subtracted from 17.1 to get 2.051?
Solution:
17.1- x = 2.051
17.1 = x + 2.051
x = 17.1 – 2.051
x = 15.049
Therefore 15.049 is subtracted from 17.1 to get 15.049
Question: 9
By how much should 34.79 be increased to get 70.15?
Solution:
34.79 + x = 70.15
x = 70.15 – 34.79
x = 35.36
Therefore 35.36 is increased to 70.15
Question: 10
By how much should 59.71 be decreased to get 34.58?
Solution:
59.71 – x = 34.58
59.71 – 34.58 = x
x = 25.13
Therefore 25.13 is decreased to get 34.58
Exercise 3.2
Question: 1
Find the product:
(i) 4.74 x 10
(ii) 0.45 x 10
(iii) 0.0215 x 10
(iv) 0.0054 x 100
Solution:
(i) 4.74 x 10
Shifting the decimal point by one place to the right
4.74 x 10 = 47.4
(ii) 0.45 x 10
Shifting the decimal point by one place to the right
0.45 × 10 = 4.5
(iii) 0.0215 x 10
Shifting the decimal point by one place to the right
0.0215 × 10 = 0.215
(iv) 0.0054 x 100
Shifting the decimal point by two places to the right
0.0054 × 100 = 0.054
Question: 2
Find the product:
(i) 35.853 x 100
(ii) 42.5 x 100
(iii) 12.075 x 100
(iv) 100 x 0.005
Solution:
(i) 35.853 x 100
Shifting the decimal point by two places to the right
35.853 = 3585.3
(ii) 42.5 x 100
Shifting the decimal point by two places to the right
42.5 × 100 = 4250
(iii) 12.075 x 100
Shifting the decimal point by two places to the right
12.075 × 100 = 1207.5
(iv) 100 x 0.005
Shifting the decimal point by two places to the right
0.005 × 100 = 0.5
Question: 3
Find the product:
(i) 2.506 x 1000
(ii) 20.708 x 1000
(iii) 0.0529 x 1000
(iv) 1000 x 0.1
Solution:
(i) 2.506 x 1000
Shifting the decimal point by three places to the right
2.506 × 1000 = 2506
(ii) 20.708 x 1000
Shifting the decimal point by three places to the right
20.708 × 1000 = 20708
(iii) 0.0529 x 1000
Shifting the decimal point by three places to the right
0.0529 × 1000 = 52.9
(iv) 1000 x 0.1
Shifting the decimal point by three places to the right
0.1 × 1000 = 100
Question: 4
Find the product:
(i) 3.4 x 17
(ii) 0.745 x 12
(iii) 28.73 x 47
(iv) 0.0415 x 59
Solution:
(i) 3.4 x 17
Multiply the number without looking into the decimal points
3.4 × 17 = 578
Mark the decimal point in the product to have one place of decimal as there in the given decimal
= 57.8
(ii) 0.745 x 12
Multiply the number without looking into the decimal points
745×12 = 8940
Mark the decimal point in the product to have three places of decimal as there in the given decimal
0.745 x 12= 8.940
(iii) 28.73 x 47
Multiply the number without looking into the decimal points
2873 × 47 = 135031
Mark the decimal point in the product to have two places of decimal as there in the given decimal
28.73 × 47 = 1350.31
(iv) 0.0415 x 59
Multiply the number without looking into the decimal points
415 × 59 = 24485
Mark the decimal point in the product to have two places of decimal as there in the given decimal
0.0415 × 59 = 2.4485
Question: 5
Find:
(i) 1.07 x 0.02
(ii) 211.9 x 1.13
(iii) 10.05 x 1.05
(iv) 13.01 x 5.01
Solution:
(i) 1.07 x 0.02
Multiply the number without looking into the decimal points
107× 2 = 214
Sum of the decimal places in the given decimals is 2 + 2 = 4
Mark the decimal point in the product to have four places of decimals
1.07 × 0.02 = 0.0214
(ii) 211.9 x 1.13
Multiply the number without looking into the decimal points
2119 × 113 = 239447
Sum of the decimal places in the given decimals is 1+ 2 = 3
Mark the decimal point in the product to have three places of decimals
211.9 × 1.13 = 239.447
(iii) 10.05 x 1.05
Multiply the number without looking into the decimal points
1005 × 105 = 105525
Sum of the decimal places in the given decimals is 2 + 2 = 4
Mark the decimal point in the product to have four places of decimals
10.05 × 1.05 = 10.5525
(iv) 13.01 x 5.01
Multiply the number without looking into the decimal points
1301 × 501 = 651801
Sum of the decimal places in the given decimals is 2+ 2 = 4
Mark the decimal point in the product to have four places of decimals
13.01 × 5.01 = 65.1801
Question: 6
Find the area of a rectangle whose length is 5.5 m and breadth is 3.4 m.
Solution:
We have
Length of rectangle = 5.5 m
Breadth of rectangle = 3.4 m
Area of rectangle = Length × Breadth
= 5.5 x 3.4
=18.7 m2
Question: 7
If the cost of a book is Rs 25.75, find the cost of 24 such books.
Solution:
Cost of one book = Rs.25.75
Therefore cost of 24 books = 25.75 x 24
= Rs.618.00
Question: 8
A car covers a distance of 14.75 km in one litre of petrol. How much distance will it cover in 15.5 litres of petrol?
Solution:
We have,
Distance covered in one litre of petrol = 14.75 km
Distance covered in 15.5 litres of petrol = 14.75 x 15.5
= 228.625 km
Question: 9
One kg of rice costs Rs 42.65. What will be the cost of 18.25 kg of rice?
Solution:
Cost of one kg rice = 42.65
Cost of 18.25kg = 42.65 x 18.25
= Rs.778.3625
Question: 10
One metre of cloth costs Rs 152.50. What is the cost of 10.75 metres of cloth?
Solution:
We have,
One metre of cloth cost = Rs.152.50
Cost of 10.75 metres =10.75 x 152.50
= Rs.1639.375
Exercise 3.3
Question: 1
Divide:
(i) 142.45 by 10
(ii) 54.25 by 10
(iii) 3.45 by 10
(iv) 0.57 by 10
(v) 0.043 by 10
(vi) 0.004 by 10
Solution:
(i) 142.45 by 10
Shifting the decimal point by one place to the left
142.45/10
= 14.245
(ii) 54.25 by 10
Shifting the decimal point by one place to the left
54.25/10
= 5.425
(iii) 3.45 by 10
Shifting the decimal point by one place to the left
3.45/10
0.345
(iv) 0.57 by 10
Shifting the decimal point by one place to the left
0.57/10
= 0.057
(v) 0.043 by 10
Shifting the decimal point by one place to the left
0.043/10
= 0.0043
(vi) 0.004 by 10
Shifting the decimal point by one place to the left
0.004/10
= 0.0004
Question: 2
Divide:
(i) 459.5 by 100
(ii) 74.3 by 100
(iii) 5.8 by 100
(iv) 0.7 by 100
(v) 0.48 by 100
(vi) 0.03 by 100
Solution:
(i) 459.5 by 100
Shifting the decimal point by two places to the left
459.5/100
= 4.595
(ii) 74.3 by 100
Shifting the decimal point by two places to the left
74.3/100
= 0.743
(iii) 5.8 by 100
Shifting the decimal point by two places to the left
5.8/100
= 0.058
(iv) 0.7 by 100
Shifting the decimal point by two places to the left
0.7/100
= 0.007
(v) 0.48 by 100
Shifting the decimal point by two places to the left
0.48/100
= 0.0048
(vi) 0.03 by 100
Shifting the decimal point by two places to the left
0.03/100
= 0.0003
Question: 3
Divide:
(i) 235.41 by 1000
(ii) 29.5 by 1000
(iii) 3.8 by 1000
(iv) 0.7 by 1000
Solution:
(i) 235.41 by 1000
Shifting the decimal point by three places to the left
235.41/1000
= 0.23541
(ii) 29.5 by 1000
Shifting the decimal point by three places to the left
29.51/000
= 0.0295
(iii) 3.8 by 1000
Shifting the decimal point by three places to the left
3.8/1000
= 0.0038
(iv) 0.7 by 1000
Shifting the decimal point by three places to the left
0.7/1000
= 0.007
Question: 4
Divide:
(i) 0.45 by 9
(ii) 217.44 by 18
(iii) 319.2 by 2.28
(iv) 40.32 by 9.6
(v) 0.765 by 0.9
(vi) 0.768 by 1.6
Solution:
(i) 0.45 by 9
= 0.45/9
= 0.05
(ii) 217.44 by 18
= 217.44/18
= 12.08
(iii) 319.2 by 2.28
= 319.2/2.28
(iv) 40.32 by 9.6
= 40.32/9.6
= 403.2/96
= 4.2
= 403.2/96 = 4.2
(v) 0.765 by 0.9
= 0.765/0.9
= 7.65/9
= 0.85
= 7.65/0.9 = 0.85
(vi) 0.768 by 1.6
= 0.768/1.6
= 7.68/16
= 0.48
Question: 5
Divide: (i) 16.64 by 20 (ii) 0.192 by 12 (iii) 163.44 by 24 (iv) 403.2 by 96 (v) 16.344 by 12 (vi) 31.92 by 228
Solution:
(i) 16.64 by 20
= 16.64/20
= 1.664/2
= 0.832
(ii) 0.192 by 12
= 0.192/12
= 0.016
(iii) 163.44 by 24
= 163.44/24
= 6.81
(iv) 403.2 by 96
= 403.2/96
= 4.2
(v) 16.344 by 12
= 16.344/12
= 1.362
(vi) 31.92 by 228
= 31.92/228
= 0.14
Question: 6
Divide:
(i) 15.68 by 20 (ii) 164.6 by 200 (iii) 403.80 by 30
Solution:
(i) 15.68 by 20
=15.68/20
=1.568/2
= 0.784
(ii) 164.6 by 200
Question: 7
(i) 76 by 0.019
(ii) 88 by 0.08
(iii) 148 by 0.074
(iv) 7 by 0.014
(iii) 403.80 by 30
Solution:
(i) 76 by 0.019
= 76/0.019
= 76000/19
= 4000
(ii) 88 by 0.08
= 88/0.08
= 8800/8
=1100
(iii) 148 by 0.074
= 148/0.074
= 148000/74
= 2000
(iv) 7 by 0.014
= 70.014
= 7000/14
= 0.823
(iii) 403.80 by 30
= 403.80/30
Question: 8
Divide:
(i) 20 by 50
(ii) 8 by 100
(iii) 72 by 576
(iv) 144 by 15
Solution:
(i) 20 by 50 = 20/50 = 0.4 (ii) 8 by 100 = 8/100 By shifting the decimal point to the left = 8/100 = 0.08 (iii) 72 by 576 = 72/576 = 0.125
(iv)144 by 15 = 144/15
= 9.6
Question: 9
A vehicle covers a distance of 43.2 km in 2.4 litres of petrol. How much distance will it travel in 1 litre of petrol?
Solution:
Distance covered in 2.4 litres of petrol = 43.2 km
Distance covered in 1 litre of petrol = 43.2/2.4
= 18km
The distance travelled in 1 litre of petrol is 18 km
Question: 10
The total weight of some bags of wheat is 1743 kg. If each bag weights 49.8 kg, how many bags are there?
Solution:
Total weight of bags of wheat = 1743 kg
Each bag weight = 49.8 kg
No of bags = 1743/49.8
= 35
Therefore the total numbers of bags are 35
Question: 11
Shikha cuts 50 m of cloth into pieces of 1.25 m each. How many pieces does she get?
Solution:
Total length of cloth = 50 m
Length of each piece of cloth = 1.25 m
Number of pieces = 50/1.25
= 5000/125
= 40 pieces
Therefore Shikha got 40 pieces
Question: 12
Each side of a rectangular polygon is 2.5 cm in length. The perimeter of the polygon is 12.5 cm. How many sides does the polygon have?
Solution:
Length of each side of rectangular polygon = 5.2 cm Perimeter of polygon = 12.5 cm No of sides polygon has = 12.5 cm No of sides polygon have = 12.5/2.5
= 5
Therefore the sides of the polygon is 5
Question: 13
The product of two decimals is 42.987. If one of them is 12.46, find the other.
Solution:
We have,
The product of the given decimals = 42.987
one decimal = 12.46
The other decimal = 42.987/12.46
= 3.45
The number is 3.45
Question: 14
The weight of 34 bags of sugar is 3483.3 kg. If all bags weigh equally, find the weight of each bag.
Solution:
Total weight of sugar = 3483.3kg
No of bags = 34
Weight of each bag = 3483.3/34
= 102.45 kg
Therefore weight of each bag is 102.45 kg
Question: 15
How many buckets of equal capacity can be filled from 586.5 litres of water, if each bucket has capacity of 8.5 litres?
Compare the following fractions by using the symbol > or < or =;
Solution:
(i) We have,
Taking the LCM of 9 and 13, we get,
9 x 13 = 117
Now, we convert the given fractions to equivalent fractions by making the denominators 117,
As we know, 91 > 72
Hence, 79 > 813
(ii) We have,
The given fractions are equivalent fractions as the denominators are equal,
And we know that, 11 > 5
Therefore, 119 > 59
(iii) We have,
Taking the LCM of 41 and 30, we get,
41 x 30 = 1230
Now, we convert the given fractions to equivalent fractions by making the denominators 1230,
Now, we clearly know 1110 > 779
Hence, 3741 > 1930
(iv) We have,
Taking the LCM of 15 and 105, we get,
5 x 3 x 7 = 105
Now, we convert the given fractions to equivalent fractions by making the denominators 105,
Question: 2
Arrange the following fractions in ascending order:
Solution:
(i) We have,
Taking the LCM of 8, 6, 8, 4 and 3, we get,
2 x 4 x 3 = 24
Now, we convert the given fractions to equivalent fractions by making the denominators 24,
We know that, 8 < 9 < 12 < 18 < 20
Hence, 13 < 38 < 24 < 68 < 56
(ii) We have,
Taking the LCM of 6, 8, 12 and 16, we get,
2 x 2 x 2 x 2 x 3 = 48
Now, we convert the given fractions to equivalent fractions by making the denominators 48,
We know that, 12 < 15 < 18 < 32
Question: 3
Arrange the following fractions in descending order:
Solution:
(i) We have,
Taking the LCM of 5, 10, 15 and 20, we get,
5 x 2 x 2 x 3 = 60
Now, we convert the given fractions to equivalent fractions by making the denominators 48
As we know 51 > 48 > 44 > 42
(ii) We have,
Taking the LCM of 7, 35, 14 and 28, we get,
7 x 5 x 2 x 2 = 140
Now, we convert the given fractions to equivalent fractions by making the denominators 140
As we know 40 > 44 > 65 > 90
Question: 4
Write the equivalent fractions of 3/5
Solution:
Multiplying or dividing both the numerator and denominator by the same number,
so that the fraction keeps its value.
So the equivalent fractions of 3/5 are
are the five equivalent fractions of 3/5
Question: 5
Find the sum:
Solution:
(i) We have,
Taking the LCM of 8 and 10, we get,
2 x 4 x 5 = 40
Now, we convert the given fractions to equivalent fractions by making the denominators 40
(ii) We have,
Taking out the LCM of 4 and 5, we get,
4 x 5 = 20
Now, we convert the given fractions to equivalent fractions by making the denominators 20
(iii) We have,
Taking out the LCM of 6 and 4, we get,
2 x 2 x 3 = 12
Now, we convert the given fractions to equivalent fractions by making the denominators 12
(iv) We have,
Taking out the LCM of 5, 10 and 15, we get,
5 x 2 x 3 = 30
Now, we convert the given fractions to equivalent fractions by making the denominators 30
Question: 6
Find the difference of
Solution:
(i) We have,
Taking out the LCM of 24 and 16, we get,
2 x 2 x 2 x 2 x 3 = 48
Now, we convert the given fractions to equivalent fractions by making the denominators 48
(ii) We have, 6 and 23/3
The difference between 6 and 23/3
(iii) We have,
Taking out the LCM of 25 and 20, we get,
5 x 5 x 4 = 100
Now, we convert the given fractions to equivalent fractions by making the denominators 100
The difference between both the fractions are
(iv) We have,
Taking out the LCM of 10 and 15, we get,
2 x 3 x 5 = 30
Now, we convert the given fractions to equivalent fractions by making the denominators 30
The difference between both the fractions are
Question: 7
Find the difference:
Solution:
(i) We have,
Taking out the LCM of 7 and 11, we get,
7 x 11 = 77
Now, we convert the given fractions to equivalent fractions by making the denominators 77
The difference between both the fractions are
(ii) We have,
(iii) We have,
(iv) We have,
Taking out the LCM of 10 and 15, we get,
2 x 3 x 5 = 30
Now, we convert the given fractions to equivalent fractions by making the denominators 30
Question: 8
Simplify:
Solution:
(i) We have,
Taking out the LCM of 3, 6 and 9, we get,
3 x 3 x 2 = 18
Now, we convert the given fractions to equivalent fractions by making the denominators 18, we get,
(ii) We have,
(iii) We have,
Taking out the LCM of 6, 8 and 12, we get,
2 x 2 x 2 x 3 = 48
Now, we convert the given fractions to equivalent fractions by making the denominators 48, we get,
Question: 9
What should be added to to get 12?
Solution:
We have,
Let x be the number added to 38/7 to get 12
Therefore,
x + 387 = 12
Question: 10
What should be added to?
Solution:
Let x be the number added to 79/15 to get 63/5
Taking out the LCM of 5 and 15, we get,
3 x 5 = 15
Now, we convert the given fractions to equivalent fractions by making the denominators 15, we get,
Question: 11
Suman studies forhours daily. She devoteshours of her time for science and mathematics. How much time does she devote for other subjects?
Solution:
Given,
Suman studies forhours daily
She devoteshours of her time for science and mathematics. Let x be time she devotes for other subjects.
Taking out the LCM of 3 and 5, we get,
3 x 5 = 15
Now, we convert the given fractions to equivalent fractions by making the denominators 48, we get,
Question: 12
A piece of wire of lengthIf it is cut into two pieces in such a way that the length of one piece iswhat is the length of the other piece?
Solution:
A piece of wire of lengthone piece is
Let the length of other piece be x m.
Question: 13
A rectangular piece of paper islong andwide. Find its perimeter?
Solution:
Given,
A rectangular piece of paper islong andwide
Perimeter = 2(length + width)
Question: 14
In a “magic square”, the sum of numbers in each row, in each column and along the diagonal is same. Is this a “magic square”?
4/11
9/11
2/11
3/11
5/11
7/11
8/11
1/11
6/11
Solution:
Given,
4/11
9/11
2/11
3/11
5/11
7/11
8/11
1/11
6/11
Therefore, the sum of numbers in each row, in each column and along the diagonal is same and the sum is 15/11.
Question: 15
The cost of Mathematics book is Rsand that of science book isWhich costs more and by how much?
Solution:
Given,
The cost of Mathematics book is Rsand that of science book is
We need to compare the cost of mathematics and science book,
Taking out the LCM of 4 and 2, we get,
2 x 2 = 4
Now, we convert the given fractions to equivalent fractions by making the denominators 4, we get
= 103/4
As we know, 103 > 82
Hence, the cost of mathematics book is more than that of the cost of the science book.
Question: 16
Provide the number in the box [] and also give its simplest form in each of the following:
Solution:
Exercise 2.2
Question: 1
Multiply
Solution:
(i) We have,
(ii) We have,
(iii) We have,
(iv) We have,
Question: 2
Find the product:
Solution:
(i) We have,
(ii) We have,
(iv) We have,
Question: 3
Simplify:
Solution:
(i) We have,
(ii) We have,
(iii) We have,
Question: 4
Find:
Solution:
(i) We have,
(ii) We have,
(iii) We have,
Question: 5
Which is greater?
Solution:
While comparing two fractions, when the numerators of both the fractions are same, then the denominator having higher value shows the fraction has lower value.
So, 6/14 is greater.
Therefore, 1/2 of 6/7 is greater.
Question: 6
(i) 7/11 of 330
(ii) 5/9 of 108 meters
(iii) 3/7 of 42 litres
(iv) 1/12 of an hour
(v) 5/6 of an year
(vi) 3/20 of a Kg
(vii) 7/20 of a litres
(viii) 5/6 of a day
(ix) 2/7 of a week
Solution:
(i) We have,
(ii) We have,
5/9 of 108 meters
= 5/9 × 108 meters
= 5 × 12 meters
60 meters
(iii) We have,
3/7 of 42 litres
= 3/7× 42 litres
= 3 × 6 litres
=18 litres
(iv) We have,
1/12 of an hour
An hour = 60 minutes
Therefore, 1/12 × 60 minutes
= 5 minutes
(v) We have,
5/6 of an year
I Year = 12 months
Therefore,
5/6 × 12 months
= 5 × 2 months
= 10 months
(vi) We have,
3/20 of a kg
1 Kg = 1000 gms
Therefore,
3/20 × 1000 gms
= 3 × 50 gms
= 150 gms
(vii) We have,
7/20 of a litre
1 litre = 1000 ml
Therefore,
7/20 × 1000 ml
= 7 × 50 ml
= 350 ml
(viii) We have,
5/6 of a day
I day = 24 hours
Therefore,
5/6 × 24 hours
= 5 × 4 hours
= 20 hours
(ix) We have,
2/7 of a week
I week = 7 days
Therefore,
2/7 × 7 days = 2 days
Question: 7
Shikha plans 5 saplings in a row in her garden. The distance between two adjacent saplings is 3/4 m. Find the distance between first and last sapling.
Solution:
There are 4 adjacent spacing for 5 saplings.
Given, the distance between two adjacent saplings is 3/4 m.
4 adjacent spacing for 5 saplings = 3/4 × 4 = 3 m
Therefore, the distance between first and last sapling is 3 m.
Question: 8
Ravish reads 1/3 part of a book in one hour. How much part of the book will be read in 2(1/5) hours?
Solution:
Let x be the full part of book.
Given, Ravish reads 1/3 part of a book in one hour
1 hour = (1/3) x
Part of the book will he read in 2(1/5) hours
Question: 9
Lipika reads a book for 1(3/4) hours every day. She reads the entire book in 6 days. How many hours in all were required by her to read the book?
Solution:
Given,
Time taken by Lipika to read a book per day = 1(3/4) = 7/4 hours
Time taken by Lipika to read a book for 6 days
Question: 10
Find the area of a rectangular park which is 41(2/3)m long and 18(3/5)m broad.
Solution:
Area of a rectangular park = (length x breadth)
Question: 11
If milk is available at Rs 17(3/4) per litre, find the cost of 7(2/5) litres of milk.
Solution:
Given,
Question: 12
Sharda can walk 8(1/3) km in one hour. How much distance will she cover in 2(2/5) hours.
Solution:
Given,
Distance covered by Sharda in one hour = 25/3 km
Question: 12
A sugar bag contains 30 kg of sugar. After consuming 2/3 of it, how much sugar is left in the bag
Solution:
Given, A sugar bag contains 30 kg of sugar
After consuming 23 of it, the amount of sugar left in the bag
Question: 14
Each side of a square is 6 (2/3)m long. Find its area.
Solution:
Given,
Question: 15
There are 45 students in a class and 3/5 of them are boys. How many girls are. there in the class?
Solution:
Given,
There are 45 students in a class,
And 3/5 of them are boys.
Therefore, no of girls in the class = 45 –3/5 × 45
= 45 – 27
= 18
Exercise 2.3
Question: 1
Find the reciprocal of each of the following fractions and classify them as proper, improper and whole numbers
(i) 3/7
(ii) 5/8
(iii) 9/7
(iv) 6/5
(v) 12/7
(vi) 1/8
Solution:
(i) 3/7
7/3 = improper number
(ii) 5/8
8/5 = improper number
(iii) 9/7
7/9 = proper number
(iv) 6/5
5/6 = proper number
(v) 12/7
7/12 = proper number
(vi) 1/8
8 = whole number
Question: 2
Divide:
Solution:
Question: 3
Divide:
Solution:
Question: 4
Simplify:
Solution:
Question: 5
A wire of length 12(1/2)m is cut into 10 pieces of equal length. Find the length of each piece.
Solution:
Question: 6
The length of a rectangular plot of area 65(1/3) m2 is 12(1/4) m. What is the width of the plot?
Solution:
Given,
The length of a rectangular plot of area 65(1/3) m2 is 12(1/4) m.
Question: 7
By what number 6(2/9) be multiplied to get 4(4/9)?
Solution:
Given,
Let x be the number which needs to be multiplied by 56/9,
Now,
Question: 8
The product of two numbers is 25(5/6). If one of the numbers is 6(2/3), find the other?
Solution:
Given,
The product of two numbers is 25(5/6). If one of the numbers is 6(2/3),
Let the other number be x.
Question: 9
The cost of 6(1/4) kg of apples is Rs 400. At what rate per kg are the apples being sold?
Solution:
Given,
Question: 10
By selling oranges at the rate of Rs 5(1/4) per orange, a fruit seller get Rs 630. How many dozens of oranges does he sell?
Solution:
Given,
By selling oranges at the rate of Rs 5(1/4) per orange, a fruit seller get Rs 630.
5(1/4) = 2(1/4)
12 apples =1 dozen
Therefore, 120 apples = 10 dozen
Question: 11
In mid-day meal scheme 3/10 litre of milk is given to each student of a primary school. If 30 litres of milk is distributed every-day in the school, how many students are there in the school?
Solution:
Given,
3/10 litre of milk is given to each student of a primary school.
30 litres of milk is distributed everyday in the school
Number of students given 3/10 litres of milk = 1
Number of students given 1 litre of milk = 10/3
Number of students given 30 litres of milk = 10/3 × 30 = 100 Students
Question: 12
In a charity show Rs 6496 were collected by selling some tickets. If the price of each ticket was Rs 50(3/4), how many tickets were sold?
Solution:
Given,
Rs 6496 were collected by selling some tickets. If the price of each ticket was Rs 50(3/4),
Question 1. Find the curved surface area and total surface area of a cylinder, the diameter of whose base is 7 cm and height is 60 cm. Solution: Diameter of the base of cylinder = 7 cm ∴ Radius (r) = 72 cm Height (h) = 60m ∴ carved surface area = 2πh = 2 x 227 x 72 x 60cm2 = 1320cm2
Question 2. The curved surface area of a cylindrical road is 132 cm2. Find its length if the radius is 0.35 cm. Solution: Curved surface area =132 cm2 Radius (r) = 0.35 cm Let h be the length of the rod Then 2πrh = 132
Question 3. The area of the base of a right circular cylinder is 616 cm2 and its height is 2.5 cm. Find the curved surface area of the cylinder. Solution: Let r be the radius of the base of the cylinder, then Area of the base = πr2
Question 4. The circumference of the base of a cylinder is 88 cm and its height is 15 cm. Find its curved surface area and total surface area. Solution: Height of the cylinder (h) = 15 cm Circumference of the base = 88 cm Let r be the radius of the base, the circumference = 2πr ∴ 2πr = 88 cm …(i)
Question 5. A rectangular strip 25 cm x 7 cm is rotated about the longer side. Find the total surface area of the solid thus generated. Solution: Dimensions of rectangular strip = 25 cm x 7 cm By rotating the strip along longer side, a solid is formed whose radius = 7 cm and height = 25 cm
Question 6. A rectangular sheet of paper 44 cm x 20 cm, is rolled along its length to form a cylinder. Find the the total surface area of the cylinder thus generated. Solution: By rolling along length wire, we get a cylinder whose circumference of its base = 20 cm
Question 7. The radii of two cylinders are in the ratio of 2 : 3 and their heights are in the ratio 5 : 3. Calculate the ratio of their curved surface areas. Solution: Ratio in radii of two cylinders = 2:3 and ratio in their heights = 5:3
Question 8. The ratio between the curved surface area and the total surface area of a right circular cylinder is 1: 2. Prove that its height and radius are equal. Solution: Let r be the radius and h be the height of a right circular cylinder, then Curved surface area = 2πrh and total surface area = 2πrh x 2πr2 = 2πr (h + r) But their ratio is 1 : 2 Hence their radius and height are equal.
Question 9. The curved surface area of a cylinder is 1320 cm2 and its base has diameter 21 cm. Find the height of the cylinder. Solution: Curved surface area of a cylinder = 1320 cm2 Diameter of its base (d) = 21 cm 21 Radius (r) = 212 cm Let h be the height of the cylinder
Question 10. The height of a right circular, cylinder is 10.5 m. If three times the sum of the areas of its two circular faces is twice the area of the curved surface area. Find the radius of its base. Solution: Height of cylinder = 10.5 m Let r be the radius and h be the height of a right circular cylinder, then Area of its two circular faces = 2π2 and area of curved surface = 2πrh Now, according to the condition:
Question 11. Find the cost of plastering the inner surface of a well at Rs 9.50 per m2, if it is 21 m deep and diameter of its top is 6 m. Solution: Diameter of the top of a cylindrical well = 6m ∴ Radius (r) = 62 = 3 m and depth (h) = 21 m ∴ Curved surface area = 2πrh = 2 x 227 x 3 x 21 m2 = 396 m2 Rate of plastering = Rs 9.50 per m2 ∴ Total cost of plastering = Rs 9.50 x 396 = Rs 3,762
Question 12. A cylindrical vessel open at the top has diameter 20 cm and height 14 cm. Find the cost of the tin-plating it on the inside at the rate of 50 paise per hundred square centimetre. Solution: Diameter of the vessel = 20 cm
Question 13. The inner diameter of a circular well is 3.5 m. It is 10 m deep. Find the cost of plastering its inner curved surface at Rs 4 per square metre. Solution: Diameter of the well = 3.5 m
Question 14. The diameter of a roller is 84 cm and its length is T20 cm. It takes 500 complete revolutions moving once over to level a playground. What is the area of the playground ? Solution: Diameter of the roller = 84 cm
Question 15. Twenty cylindrical pillars of the Parliament House are to be cleaned. If the diameter of each pillar is 0.50 m and height is 4 m, what will be the cost of cleaning them at the rate of Rs 2.50 per square metre ? Solution: Diameter of each pillar = 0.50 m
Question 16. The total surface area of a hollow cylinder which is open from both sides is 4620 sq. cm, area of the base ring is 115.5 sq. cm and height 7 cm. Find the thickness of the cylinder. Solution: Total surface area of a hollow cylinder opened from both sides = 4620 cm2 Area of base ring = 115.5 cm2 Height of cylinder (h) = 7 cm Let R be the outer radius and r be the inner radius
Question 17. The sum of the radius of the base and height of a solid cylinder is 37 m, if the total surface area of the solid cylinder is 1628 m2, find the circumference of its base. Solution: Let r be the radius and h be the height of the solid cylinder, then r + h = 37 m …(i) Total surface area = 1628 ⇒ 2πr (r + h) = 1628
Question 18. Find the ratio between the total surface area of a cylinder to is curved surface area,given that its height and radius are 7.5 cm and 3.5 cm. Solution: Radius (r) of cylinder = 3.5 cm and height (h) = 7.5 cm ∴ Curved surface area = 2πrh and total surface area = 2πr (h + r) ∴ Ratio = 2πr (h + r)- 2πrh = (h + r): h = 7.5 + 3.5 : 7.5 ⇒ 11 : 7.5
Question 19. A cylindrical vessel, without lid, has to be tin-coated on its both sides. If the radius of the base is 70 cm and its height is 1.4 m, calculate the cost of tin-coating at the rate of Rs 3.50 per 1000 cm2. Solution: Radius of the vessel (r) = 70 cm and height (h) = 1.4 m = 140 cm ∴ Area of inner and outer curved surfaces and bases = 2 x 2πrh + 2πr2
Exercise 22.2
Question 1. Find the volume of a cuboid whose (i) r = 3.5 cm, h = 40 cm (ii) r = 2.8 m, h = 15 m Solution: (i) Radius (r) = 3.5 cm Height (h) = 40 cm Volume of cylinder = πr2h
Question 2. Find the volume of a cylinder, if the diameter (d) of its base and its altitude (h) are: (i) d= 21 cm, h = 10 cm (ii) d = 7 m, h = 24 m. Solution: (i) Diameter (d) = 21 cm
Question 3. The area of the base of a right circular cylinder is 616 cm3 and its height is 25 cm. Find the volume of the cylinder. Solution: Base area of cylinder = 616 cm2 Height (h) = 25 cm. ∴ Volume = Area of base x height = 616 x 25 cm3 = 15400 cm3
Question 4. The circumference of the base of a cylinder is 88 cm and its height is 15 cm. Find the volume of the cylinder. Solution: Circumference of the base of cylinder = 88 cm Let r be the radius
Question 5. A hollow cylindrical pipe is 21 dm long. Its outer and inner diameters are 10 cm and 6 cm respectively. Find the volume of the copper used in making the pipe. Solution: Length (Height) of hollow cylindrical pipe = 21 dm = 210 cm Inner diameter = 6 cm Outer diameter = 10 cm
Question 6. Find the (i) curved surface area (ii) total surface area and (iii) volume of a right circular cylinder whose height is 15 cm and the radius of the base is 7 cm Solution: Radius of the cylider (r) = 7 cm and height (h) = 15 cm (i) Curved surface area = 2πrh
Question 7. The diameter of the base of a right circular cylinder is 42 cm and its height is 10 cm. Find the volume of the cylinder. Solution: Diameter of the base of cylinder = 42 cm
Question 8. Find the volume of a cylinder, the diameter of whose base is 7 cm and height being 60 cm. Also, find the capacity of the cylinder in litres. Solution: Diameter of cylinder = 7 cm
Question 9. A rectangular strip 25 cm x 7 cm is rotated about the longer side. Find the volume of the solid, thus generated. Solution: Size of rectangular strip = 25 cm x 7 cm By rotating, about the longer side, we find a right circular cylinder, then the radius of cylinder (r) = 7 cm
Question 10. A rectangular sheet of paper 44 cm x 20 cm, is rolled along its length to form a cylinder. Find the volume of the cylinder so formed. Solution: Size of sheet of paper = 44 cm x 20 cm By rolling along length, a cylinder is formed in which circumference of its base = 20 cm and height (h) = 44 cm
Question 11. The volume and the curved surface area of a cylinder are 1650 cm3 and 660 cm2 respectively. Find the radius and height of the cylinder. Solution: Volume of cylinder = 1650 cm3 and curved surface area = 660 cm2 Let r be the radius and h be the height of the cylinder, then
Question 12. The radii of two cylinders are in the ratio 2 :3 and their heights are in the ratio 5 : 3. Calculate the ratio of their volumes. Solution: Ratio in radii = 2:3 and in heights = 5:3 Let r1,h1 and r2, h2 are the radii and heights of two cylinders respectively.
Question 13. The ratio between the curved surface area and the total surface area of a right circular cylinder is 1 : 2. Find the volume of the cylinder, if its total surface area is 616 cm2. Solution: Ratio in curved surface are and total surface area =1 : 2 Let r be the radius and h be the height. Then 2πrh : 2πr (h + r)= 1 : 2
Question 14. The curved surface area of a cylinder is 1320 cm2 and its base has diameter 21 cm. Find the volume of the cylinder. Solution: Curved surface area of a cylinder = 1320 cm2 Diameter of base = 21 cm
Question 15. The ratio between the radius of the base and the height of a cylinder is 2 : 3. Find the total surface area of the cylinder, if its volume is 1617 cm3. Solution: Ratio between radius and height of a cylinder = 2:3 Volume = 1617 cm3 Let r be the radius and A be the height of the cylinder, then
Question 16. The curved surface area of a cylindrical pillar is 264 m2 and its volume is 924 m3. Find the diameter and the height of the pillar. Solution: Let r be the radius and A be the height of the cylinder, then 2πrh = 264 …(i) and πr2h= 924 …(ii) Dividing (ii) by (i),
Question 17. Two circular cylinders of equal volumes have their heights in the ratio 1 : 2. Find the ratio of their radii. Solution: Volumes of two cylinders are equal. Let r1, r2 are the radii and h1, h2 are their heights, then Now, volume of first cylinder = πr12h1 and Volume of second cylinder = πr22h2 Their volumes are equal
Question 18. The height of a right circular cylinder is 10.5 m. Three times the sum of the areas of its two circular faces is twice the area of the curved surface. Find the volume of the cylinder. Solution: Height of cylinder (A) = 10.5 m. Let r be the radius, then Sum of areas of two circular faces = 2π2 and curved surface area = 2πrh According to the condition,
Question 19. How many cubic metres of earth must be dug-out to sink a well 21 m deep and 6 m diameter ? Solution: Diameter of well = 6 m
Question 20. The trunk of a tree is cylindrical and its circumference is 176 cm. If the length of the trunk is 3 m, find the volume of the timber that can be obtained from the trunk. Solution: Circumference of cylindrical trunk = 176 cm
Question 21. A well is dug 20 m deep and it has a diameter of 7 m. The earth which is so dug out is spread out on a rectangular plot 22 m long and 14 m broad, what is the height of the platform so formed ? Solution: Diameter of the well = 7 m ∴ Radius (r) = 72 m Depth (h) = 20 m = Volume of earth dug out = πr2h
Question 22. A well with 14 m diameter is dug 8 m deep. The earth taken out of it has been evenly spread all around it to a width of 21 m to form an embankment. Find the height of the embankment. Solution: Diameter of the well = 14 m
Question 23. A cylindrical container with diameter of base 56 cm contains sufficient water to submerge a rectangular solid of iron with dimensions 32 cm x 22 cm x 14 cm. Find the rise in the level of the water when the solid is completely submerged. Solution: Diameter of the base of a cylindrical container = 56 cm Dimensions of the rectangular solid = 32 cm x 22 cm x 14 cm ∴ Volume of solid = 32 x 22 x 14 cm3 = 9856 cm3 ∴Volume of water rose up = 9856 cm3 Let h be the height of water, then πr2h = 9856
Question 24. A rectangular sheet of paper 30 cm x 18 cm can be transformed into the curved surface of a right circular cylinder in two ways Le., either by rolling the paper along its length or by rolling it along its breadth. Find the ratio of the volumes of the two cylinders thus formed. Solution: Size of paper = 30 cm and 18 cm. (i) By rolling length wise, The circumference of base = 30 cm
Question 25. The rain which falls on a roof 18 m long and 16.5 m wide is allowed to be stored in a cylindrical tank 8 m in diameter. If it rains 10 cm on a day, what is the rise of water level in the tank due to it. Solution: Length of roof (l) = 18 m Breadth (b) = 16.5 m Height of water on the roof = 10 cm ∴ Volume of water collected
Question 26. A Piece of ductile metal is in the form of a cylinder of diameter 1 cm and length 5 cm. It is drawnout into a wire of diameter 1 mm. What will be the length of the wire so formed ? Solution: Diameter of ductile metal = 1 cm and length (h) = 5 cm and radius (r) = 12 cm
Question 27. Find the length of 13.2 kg of copper wire of diameter 4 mm when 1 cubic cm of copper weighs 8.4 gm. Solution: Weight of wire = 13.2 kg Diameter of wire = 4 mm
Question 28. 2.2 cubic dm of brass is to be drawn into a cylindrical wire 0.25 cm in diameter. Find the length of the wire. Solution: Volume of brass = 2.2 cu.dm = 2.2 x 10 x 10 x 10 = 2200 cm2 Diameter of wire = 0.25 cm
Question 29. The difference between inside and outside surfaces of a cylindrical tube 14 cm long is 88 sq. cm. If the volume of the tube is 176 cubic cm, find the inner and outer radii of the tube. Solution: Let r and R be the radii of inner and outer surfaces of a cylindrical tube,
Question 30. Water flows out through a circular pipe whose internal diameter is 2 cm, at the rate of 6 metres per second into a cylindrical tank, the radius of whose base is 60 cm. Find the rise in the level of water in 30 minutes. Solution: Diameter of pipe = 2 cm ∴ Radius (r) = 22 = 1 cm = 0.01 m Length of flow of water in 1 second = 6 m Length of flow in 30 minutes = 6 x 30 x 60 m = 10800 m ∴ Volume of water = πr2h
Question 31. A cylindrical tube, open at both ends, is to made of metal. The internal diameter of the tube is 10.4 cm and its length is 25 cm. The thickness of the metal tube is 8 mm everywhere. Calculate the volume of the metal. Solution: Length of metal tube (h) = 25 cm Internal diameter = 10.4 cm ∴ Internal radius (r) = 10.42 = 5.2 cm Thickness of metal = 8 mm = 0.8 cm ∴ Outer radius = 5.2 + 0.8 = 6 cm Now volume of the metal
Question 32. From a tap of inner radius 0.75 cm, water flows at the rate of 7 m per second. Find the volume in litres of water delivered by the pipe in one hour. Solution: Inner radius of pipe (r) = 0.75 cm Rate of water flow = 7 m per second ∴ Length of water flow in 1 hour (h) = 7 x 3600 m = 25200 m ∴ Volume of water in 1 hour
Question 33. A cylindrical water tank of diameter 1. 4 m and height 2.1 m is being fed by a pipe of diameter 3.5 cm through which water flows at the rate of 2 metre per second. In how much time the tank will be filled ? Solution: Diameter of cylindrical tank = 1.4 m ∴ Radius (r) = 1.42 = 0.7 m Height (h) = 2.1m.
Question 34. A rectangular sheet of paper 30 cm x 18 cm can be transformed into the curved surface of a right circular cylinder in two ways Le. either by rolling the paper along its length or by rolling it along its breadth. Find the ratio of the volumes of the two cylinder thus formed. Solution: In first case, By rolling the paper along its length, the circumference of the base = 30 cm and height (h) = 18 cm Note: See Q.No. 24 this exercise
Question 35. How many litres of water flow out of a pipe having an area of cross-section of 5 cm2 in one minute, if the speed of water in the pipe is 30 cm/sec ? Solution: Speed of water = 30 cm/sec ∴ Water flow in 1 minute = 30 cm x 60 = 1800 cm Area of cross-section = 5 cm2 ∴ Volume of water = 1800 x 5 = 9000 cm3 Capacity of water in litres = 9000 x l ml
Question 36. A solid cylinder has a total surface area of 231 cm2. Its curved surface area is 23 of the total surface area. Find the volume of the cylinder. Solution: Total surface area of a solid cylinder = 231 cm2 Curved surface area = 23 of 231 cm2 = 2 x 77 = 154 cm2 and area of two circular faces = 231-154 = 77 cm2 Let r be the radius, then
Question 37. Find the cost of sinking a tubewell 280 m deep, having diameter 3 m at the rate of Rs 3.60 per cubic metre. Find also the cost of cementing its inner curved surface at Rs 2.50 per square metre. Solution: Diameter of well = 3 m ∴ Radius (r) = 32 m and depth (h) = 280 m (i) Volume of earth dug out = πr2h Rate of sinking the well = Rs 3.60 per m3 ∴ Total cost of sinking = Rs 1980 x 3.60 = Rs 7,128 (ii) Inner curved surface area = 2πrh Rate of cementing = Rs 2.50 per m2 ∴ Total cost of cementing = Rs 2.50 x 2640 = Rs 6,600
Question 38. Find the length of 13.2 kg of copper wire of diameter 4 mm, when 1 cubic cm of copper weighs 8.4 gm. Solution: Note : See Q.No. 27 of this exercise
Question 39. 2.2 cubic dm of brass is to be drawn into a cylindrical wire 0.25 cm in diameter. Find the length of the wire. Solution: Note: See Q.No. 28 of this exercise.
Question 40. A well with 10 m inside diameter is dug 8.4 m deep. Earth taken out of it is spread all around it to a width of 7.5 m to form an embankment. Find the height of the embankment. Solution: Diameter of the well = 10 m
Question 41. A hollow garden roller, 63 cm wide with a girth of 440 cm, is made of 4 cm thick iron. Find the volume of the iron. Solution: Width of roller (h) = 63 cm. Outer circumference of roller = 440 cm
Question 42. What length of a solid cylinder 2 cm in diameter must be taken to recast into a hollow cylinder of length 16 cm, external diameter 20 cm and thickness 2.5 mm ? Solution: Length of hollow cylinder (h) = 16 cm External diameter = 20 cm ∴ External radius (R) = 202 = 10 cm Thickness of iron = 2.5 mm ∴ Internal radius (r) = 10 – 0.25 = 9.75 cm ∴ Volume of iron = πh (R2 – r2)
Question 43. In the middle of rectangular field measuring 30 m x 20 m, a well of 7 m diameter and 10 m depth is dug. The earth so removed is evenly spread over the remaining part of the field. Find the height through which the level of the field is raised. Solution: Diameter of well = 7 m ∴ Radius (r) = 72 m Depth (h) = 10 m. ∴ Volume of earth dug out = πr2h
Question 1. Find the volume of a cuboid whose (i) length = 12 cm, breadth = 8 cm, height = 6 cm (ii) length = 1.2 m, breadth = 30 cm, height = 15 cm (iii) length = 15 cm, breadth = 2.5 dm, height = 8 cm Solution: In a cuboid, (i) Length (l) = 12 cm Breadth (b) = 8 cm Height (h) = 6 cm ∴ Volume = Ibh = 12 x 8 x 6 cm3 = 576 cm3 (ii) Length (l) = 1.2 m = 120 cm breadth (6) = 30 cm Height (h) = 15 cm ∴ Volume = Ibh = 120 x 30 x 15 cm3 = 54000 cm3 (iii) Length (l) = 15 cm Breadth (b) = 2.5 dm = 25 cm Height (h) = 8 cm ∴ Volume = Ibh = 15 x 25 x 8 cm3 = 3000 cm2
Question 2. Find the volume of the cube whose side is (i) 4 cm (ii) 8 cm (iii) 1.5 dm (iv) 1.2 m (v) 25 mm. Solution: (i) Side of a cube (a) = 4 cm ∴ Volume = a3 = (4)3 cm3 = 4 x 4 x 4 = 64 cm3 (ii) Side of cube (a) = 8 cm ∴ Volume = a3 = (8)3 4 cm = 8 x 8 x 8 cm3 = 512 cm3 (iii) Side of cube (a) = 1.5 dm = 15 cm ∴ Volume = a3 = (1.5)3 dm2 = (15)3 cm3 = 15 x 15 x 15 = 3375 cm3 (iv) Side of cube (a) = 1.2 m = 120 cm ∴ Volume = a3 = (120)3 cm3 = 120 x 120 x 120 = 1728000 cm3 (v) Side of cube (a) = 25 mm = 2.5 cm. ∴ Volume = a3 = (2.5)3 cm3 = 2.5 x 2.5 x 2.5 cm3 = 15.625 cm3
Question 3. Find the height of a cuboid of volume 100 cm3 whose length and breadth are 5 cm and 4 cm respectively. Solution: Volume of a cuboid =100 cm3 Length (1) = 5 cm and breadth (b) = 4 cm
Question 4. A cuboidal vessel is 10 cm long and 5 cm wide, how high it must be made to hold 300 cm3 of a liquid ? Solution: Volume of the liquid in the vessel = 300 cm3 Length (l)= 10 cm Breadth (b) = 5 cm
Question 5. A milk container is 8 cm long and 50 cm wide. What should be its height so that it can hold 4 litres of milk ? Solution: Capacity of milk = 4 litres ∴ Volume of the container = 4 x 1000 cm3 = 4000 cm3 Length (l) = 8 cm Width (b) = 50 cm
Question 6. A cuboidal wooden block contains 36 cm3 wood. If it be 4 cm long and 3 cm wide, find its height. Solution: Volume of wooden cuboid block = 36 cm3 Length (l) = 4 cm Breadth (b) = 3 cm
Question 7. What will happen to the volume of a cube, if its edge is (i) halved (ii) trebled ? Solution: Let side of original cube = a cm ∴ Volume = a3 cm3 (i) In first case, (ii) In second case, when side (edge) is trebled, then side = 3a ∴ Volume = (3a)3 = 27a3 ∴ It will be 27 times
Question 8. What will happen to the volume of a cuboid if its (i) Length is doubled, height is same and breadth is halved ? (ii) Length is doubled, height is doubled and breadth is same ? Solution: Let l, b and h be the length, breadth and height of the given cuboid respectively. ∴ Volume = lbh. (i) Length is doubled = 21 ∴ The volume will be the same. (ii) Length is doubled = 21 breadth is same = b height is doubled = 2h ∴ Volume = 2l x b x 2h = 4 lbh ∴ Volume will be 4 times
Question 9. Three cuboids of dimensions 5 cm x 6 cm x 7 cm, 4 cm x 7 cm * 8 cm and 2 cm x 3 cm x 13 cm are melted and a cube is made. Find the side of cube. Solution: Dimensions of first cuboid = 5 cm x 6 cm x 7 cm ∴ Volume = 5 x 6 x 7 = 210 cm3 Dimensions of second cuboid = 4 cm x 7 cm x 8 cm ∴ Volume = 4x 7 x 8 = 224 cm3 Dimensions of third cuboid = 2 cm x 3 cm x 13 cm ∴ Volume = 2 x 3 x 13 = 78 cm3 Total volume of three cubes = 210 + 224 + 78 cm3 = 512 cm3 ∴ Volume of cube = 512 cm3
Question 10. Find the weight of solid rectangular iron piece of size 50 cm x 40 cm x 10 cm, if 1 cm3 of iron weighs 8 gm. Solution: Dimension of cuboidal iron piece = 50 cm x 40 cm x 10 cm ∴ Volume = 50 x 40 x 10 = 20000 cm3 Weight of 1 cm3 = 8 gm ∴ Total weight of piece = 20000 x 8 gm
Question 11. How many wooden cubical blocks of side 25 cm can be cut from a log of wood of size 3 m by 75 cm by 50 cm, assuming that there is no wastage ? Solution: Length of log (l) = 3 m = 300 cm. Breadth (b) = 75 cm and height (h) = 50 cm ∴ Volume of log = lbh = 300 x 75 x 50 cm3 = 1125000 cm3 Side of cubical block = 25 cm ∴ Volume of one block = a2 = 25 x 25 x 25 cm3 = 15625 cm3 ∴ Number of blocks to be cut out
Question 12. A cuboidal block of silver is 9 cm long, 4 cm broad and 3.5 cm in height. From it, beads of volume 1.5 cm2 each are to be made. Find the number of beads that can be made from the block ? Solution: Length of block (l) = 9 cm Breadth (b) = 4 cm and height (h) = 3.5 cm ∴ Volume = l x b x h = 9 x 4 x 3.5 cm3 = 126 cm3 Volume of one bead = 1.5 cm3 ∴ Number of beads = 126105 = 84
Question 13. Find the number of cuboidal boxes measuring 2 cm by 3 cm by 10 cm which can be stored in a carton whose dimensions are 40 cm, 36 cm and 24 cm. Solution: Length of cuboidal box (l) = 2 cm breadth (b) = 3 cm and height (h) = 10 cm ∴ Volume = lx b x h = 2 x 3 x 10 = 60 cm3 Volume of carton = 40 x 36 x 24 cm3 = 34560 cm3 ∴ Number of boxes to be height in the carton
Question 14. A cuboidal block of solid iron has dimensions 50 cm, 45 cm and 34 cm. How many cuboids of size 5 cm by 3 cm by 2 cm can be obtained from the block ? Assume cutting causes no wastage. Solution: Dimensions of block = 50 cm, 45 cm, 34 cm ∴ Volume = 50 x 45 x 34 = 76500 cm3 Size of cuboid = 5 cm x 3 cm x 2 cm ∴ Volume of cuboid = 5 x 3 x 2 = 30 cm3
Question 15. A cube A has side thrice as long as that of cube B ? What is the ratio of the volume of cube A to that of cube B ? Solution: Let side of cube B = a Then Volume = a3 and side of cube A = 3a Volume = (3a)3 = 3a x 3a x 2a = 27a3 ∴ Ratio of volume’s A and B = 27a3 : a3 = 27 : 1
Question 16. An ice-cream brick measures 20 cm by 10 cm by 7 cm. How many such bricks can be stored in a deep fridge whose inner dimensions are 100 cm by 50 cm by 42 cm ? Solution: Dimensions of ice cream brick = 20 cm x 10 cm x 7 cm ∴ Volume = 20 x 10 x 7 cm3 = 1400 cm3 Dimensions of inner of fridge = 100 cm x 50 cm x 42 cm = 210000 cm3 ∴ Number of bricks to be kept in the fridge
Question 17. Suppose that there are two cubes, having edges 2 cm and 4 cm, respectively. Find the volume V1 and V2 of the cubes and compare them. Solution: Side of first cube (a) = 2 cm ∴ Volume (V1) = a3 = (2) = 8 cm3 Similarly side of second cube = 4 cm and volume (V2) = (4)3 = 64 cm3 Now V2 = 64 cm3 = 8 x 8 cm3 = 8 x V1 ⇒ V2 = 8V1
Question 18. A tea-packet measures 10 cm x 6 cm x 4 cm.How many such tea-packets can be placed in a cardboard box of dimensions 50 cm x 30 cm x 0.2 m ? Solution: Dimensions of tea-packet = 10cm x 6cm x 4 cm ∴ Volume =10 x 6 x 4 = 240 cm3 Dimensions of box = 50 cm x 30 cm x 0.2 m = 50 cm x30 cm x20 cm ∴ Volume = 50 x 30 x 20 = 30000 cm3 ∴ Number of tea-packets to be kept = 30000240
Question 19. The weight of a metal block of size 5 cm by 4 cm by 3 cm is 1 kg. Find the weight of a block of the same metal of size 15 cm by 8 cm by 3 cm. Solution: Dimensions of a metal block = 5 cm x 4 cm x 3 cm = 5 x 4 x 3 = 60 cm3 Dimensions of a second block = 15 cm x 8 cm x 3 cm = 15 x 8 x 3 = 360 cm3 But weight of first block = 1 kg ∴ Weight of second block = 116 x 360 = 6 kg
Question 20. How many soap cakes can be placed in a box of size 56 cm x 0.4 m x 0.25 m, it the size of soap cake is 7 cm x 5 cm x 2.5 cm ? Solution: Size of box = 56 cm x 0.4 m x 0.25 m = 56 cm x 40 cm x 25 cm ∴ Volume = 56 x 40 x 25 cm3 = 56000 cm3 Size of a soap cake = 7 cm x 5 cm x 2.5 cm ∴ Volume = 7 x 5 x 2.5 cm3 = 87.5 cm3 ∴ Number of cakes to be kept in the box = 5600087.5 = 640
Question 21. The volume of a cuboid box is 48 cm3. If its height and length are 3 cm and 4 cm respectively, find its breadth. Solution: Volume of cuboid box = 48 cm3 Length (l) = 4 cm Height = (h) = 3 cm
Exercise 21.2
Question 1. Find the volume in cubic metres (cu.m) of each of the cuboids whose dimensions are : (i) length = 12 cm, breadth = 10 m, height = 4.5 m (ii) length = 4 m, breadth = 2.5 m, height = 50 cm (iii) length = 10 m, breadth = 25 dm, height = 25 cm. Solution: (i) Length of cuboid (l) = 12 m Breadth (b) = 10m and height (h) = 4.5 m ∴Volume = l x b x h = 12 x 10 x 4.5 m3 = 540 m3 (ii) Length of cuboid (l) = 4 m Breadth (b) = 2.5m Height (h) = 50 cm = 0.5 m ∴ Volume = l x b x h = 4 x 2.5 x 0.5 = 5 m3 (iii) Length of cuboid (l) = 10 m Breadth (b) = 25 dm = 2.5 m Height (h) = 25 cm 0.25 m ∴ Volume = l x b x h = 10 x 2.5 x 0.25 m3 = 6.25 m3
Question 2. Find the volume in cubic decimetre of each of the cubes whose side is (i) 1.5 m (it) 75 cm (iii) 2 dm 5 cm Solution: (i) Side of cube (a) = 1.5 m ∴ Volume = a3 = (1.5)3 m3 = 1.5 x 1.5 x 1.5 m3 = 3.375 m3 = 3.375 x 1000 = 3375 dm3 (ii) Side of cube (a) = 75 cm = 7.5 dm ∴ Volume = a3 = (7.5)3 dm3 = 421.875 dm3 (iii) Side of cube (a) = 2 dm 5 cm = 2.5 dm ∴ Volume = (a)3 = (2.5)3 dm3 = 15.625 dm3
Question 3. How much clay is dug out in digging a well measuring 3m by 2m by 5m? Solution: Length of well (l) = 3m breadth (b) = 2 m and height (depth) (h) = 5 m Volume of earth dug out = l x b x h = 3 x 2 x 5 = 30m3
Question 4. What will be the height of a cuboid of volume 168 m3, if the area of its base is 28 m2 ? Solution: Volume of a cuboid = 168 m3 Area of its base l.e., l x b = 28 m3
Question 5. A tank is 8 m long, 6 m broad and 2 m high. How much water can it contain ? Solution: Length of tank (l) = 8 m Breadth (b) = 6 m Height (h) = 2 m ∴ Volume of water in the tank = l x b x h = 8 x 6 x 2 = 96 m3 = 96 x 1000 = 96000litres (∵1m3 = 1000litre)
Question 6. The capacity of a certain cuboidal tank is 50000 litres of water. Find the breadth of the tank if its height and length are 10 m and 2.5 m respectively. Solution: Capacity of water in the tank = 50000 litres ∴ Volume of water = 50000 x 11000 = 50 m3 (1000 l = 1 m3) Height of tank (h)= 10 m and length (l) = 2.5 m Volume 50
Question 7. A rectangular diesel tanker is 2 m long, 2 m wide and 40 cm deep. How many litres of diesel can it hold ? Solution: Length of tanker (l) = 2 m Breadth (b) = 2m Depth (h) = 40 cm = 0.4 m ∴ Volume = l x bx h = 2 x 2 x 0.4=1.6m3 Quantity of diesel = 1.6 x 1000 litres (1 m3= 1000 l) = 1600 litres
Question 8. The length, breadth and height of a room are 5 m, 4.5 m and 3 m, respectively. Find the volume of the air it contains. Solution: Length of room (l) = 5 m Breadth (6) = 4.5 m and height (h) = 3 m ∴ Volume of air it contains = l x b x h = 5 x 4.5 x 3 m3 = 67.5 m3
Question 9. A water tank is 3 m long, 2 m broad and 1 m deep. How many litres of water can it hold ? Solution: Length of tank (l) = 3 m Breadth (b) = 2 m and depth (h) = 1 m ∴ Volume of tank = l x b x h = 3 x 2 x 1 = 6 m3 ∴ Quantity of water it can contains = 6 x 1000 litres = 6000 litres (1 m3= 1000 litres)
Question 10. How many planks each of which is 3 m long, 15 cm broad and 5 cm thick can be prepared from a wooden block 6 m long, 75 cm broad and 45 cm thick ? Solution: Length of wooden block (l) = 6 m Width (b) = 75 cm = 0.75 m Thickness (h) = 45 cm = 0.45 m ∴ Volume = l x b x h = 6 x 0.75 x 0.45 m3 Length of plank (l) = 3 m Breadth (b) = 15 cm = 0.15 m Thickness (h) = 5 cm = 0.05 m ∴ Volume = 3 x 0.15 x 0.05 m3 Number of planks
Question 11. How many bricks each of size 25 cm x 10 cm x 8 cm will be required to build a wall 5 m long, 3 m high and 16 cm thick assuming that the volume of sand and cement used in the construction is negligible ? Solution: Size of one brick = 25 cm x 10 cm x 8 cm ∴ Volume of one brick = 25 x 10 x 8 cm3 Length of wall (l) = 5 m Width (b) = 0.16 m Height (h) = 3 m ∴ Volume of wall = l x b x h = 5 x 0.16 x 3 m3 = 2.4 m3 ∴ Number of bricks required
Question 12. A village, having a population of 4000 requires 150 litres water per head per day. It has a tank which is 20 m long, 15 m broad and 6 m high. For how many days the water of this tank will last ? Solution: Total population of a village = 4000 Water required for each person for one day = 150 litres ∴ Water required for 4000 persons for one day = 150 x 4000 = 600000 litres Length of tank (l) = 20 m Breadth (b) = 15 m Height (h) = 6 m ∴ Volume of tank = l x b x h = 20 x 15 x 6 m3 = 1800 m3 Capacity of water in the tank = 1800 x 1000 l= 1800000l (1 m3 = 1000 l) ∴ Number of days, the water will last
Question 13. A rectangular field is 70 m long and 60 m broad. A well of dimensions 14 m x 8 m x 6 m is dug outside the field and the earth dugout from this well is spread evenly on the field. How much will the earth level rise ? Solution: Length of well (l) = 14 m Breadth (A) = 8m Depth (A) = 6m ∴ Volume of earth dugout = l x bx h = 14 x 8 x 6 = 672 m3 Length of field = 70 m and breadth = 60 m Let h be the height of earth spread over Then 70 x 60 x h = 672 ⇒ h = 67270×60 = 0.16m ∴ Height of earth = 0.16 m = 16 cm
Question 14. A swimming pool is 250 m long and 130 m wide. 3250 cubic metres of water is pumped into it. Find the rise in the level of water. Solution: Volume of water = 3250 m3 Length of pool (l) = 250 m Breadth (b)= 130 m ∴ Height of water level
Question 15. A beam 5 m long and 40 cm wide contains 0.6 cubic metres of wood. How thick is the beam? Solution: Volume of wood of the beam = 0.6 m3 = 600000 Length of beam (l) = 5 m = 500 cm Breadth (b) = 40 cm
Question 16. The rainfall on a certain day was 6 cm. How many litres of water fell on 3 hectares of field on that day ? Solution: Area of the field = 3 hectares = 3 x 10000 square metres = 30000 square metres Height of rainfall = 6 cm = m3
Question 17. An 8 m long cuboidal beam of wood when sliced produces four thousand 1 cm cubes and there is no wastage of wood in this process. If one edge of the beam is 0.5 m, find the third edge. Solution: Length of cuboidal beam (l) = 8 m = 800 cm Number of cubical sliced = 4000 Edge of each cube = 1 cm Volume of beam = 4000 (1)3 cm3 = 4000 cm3 One edge of the beam = 0.5 m = 50 cm.
Question 18. The dimensions of a metal block are 2.25 m by 1.5 m by 27 cm. It is melted and recast into cubes, each of the side 45 cm. How many cubes are formed ? Solution: Dimensions of metal block = 2.25 m x 1.5 m x 27 cm ∴ Volume = 2.25 x 1.5 x 0.27 m3 = 225 x 150 x 27 cm3 = 911250 cm3 Side of each cube (a) = 45 cm ∴ Volume of one cube = a3 = (45)3 cm3 = 91125 cm3 ∴ Number of cubes = 91125091125 = 10
Question 19. A solid rectangular piece of iron measures 6 m by 6 cm by 2 cm. Find the weight of this piece if 1 cm3 of iron weighs 8 gm. Solution: Dimensions of a piece of rectangular iron = 6m x 6cm x 2cm ∴ Volume = 600 x 6 x 2 cm3 = 7200 cm3 Weight of 1 cm3 = 8 gm ∴ Total weight of the piece = 7200 x 8 gm = 57600 gm = 576001000 kg = 57.6 kg
Question 20. Fill in the blanks in each of the following so as to make the statement true : (i) 1 m3 = ……… cm3 (ii) 1 litre = …….. cubic decimetre (iii) 1 kl = …… m3 (iv) The volume of a cube of side 8 cm is …….. . (v) The volume of wooden cuboid of length 10 cm and breadth 8 cm is 4000 cm3. The height of the cuboid is …….. cm (vi) 1 cu.dm = …….. cu.mm (vii) 1 cu.km = ……cu.m (viii) 1 litre =…….. cu.cm (ix) 1 ml = ……… cu.cm (x) 1 kl = ……… cu.dm = ……. cu.cm. Solution: (i) 1 m3 = 1000000 or 106 cm3 (ii) 1 litre = 1 cubic decimetre (iii) 1 kl = 1 m3 (iv) The volume of a cube of side 8 cm is 512 cm3 (V = a3 = 8 x 8 x 8 = 512 cm3) (v) The volume of a wooden cuboid of length 10 cm and breadth 8 cm is 4000 cm3. The height of the cuboid is 50 cm (vi) 1 cu.dm = 1000000 cu mm = 106 cu.mm (vii) 1 cu.km = 1000 x 1000 x 1000 cu.m = 109 cu.m (viii) 1 litre = 1000 cu.cm = 103 cu.cm (ix) 1 ml = 1 cu.cm (x) 1 kl = 1000 cu.dm = 100 x 100 x 100 cu.cm = 106 cu.cm
Exercise 21.3
Question 1. Find the surface area of a cuboid whose : (i) length = 10 cm, breadth = 12 cm and height = 14 cm (ii) length = 6 dm, breadth = 8 dm, height = 10 dm (iii) length = 2 m, breadth = 4 m and height = 5 m (iv) length = 3.2 m, breadth = 30 dm, height = 250 cm. Solution: (i) Length of cuboid (l) = 10 cm Breadth (b) = 12 cm Height (h) = 14 cm ∴ Surface area = 2(1 × b + b × h + h × l) = 2(10 x 12 + 12 x 14 + 14 x 10) cm2 = 2(120+ 168 + 140) cm2 = 2 x 428 = 856 cm2 (ii) Length of cuboid (l) = 6 dm Breadth (b) = 8 dm Height (h) = 10 dm ∴ Surface area = 2 ( l × b + b x h + h× l) = 2(6 x 8 + 8 x 10 + 10 x 6) dm2 = 2(48 + 80 + 60) dm2 = 2 x 188 = 376 dm2 (iii) Length of cuboid (l) = 2 m Breadth (b) = 4 m Height (h) = 5 m ∴ Surface area = 2(l × b + b × h + h × l) = 2(2 x 4 + 4 x 5 + 5 x 2) m2 = 2(8 + 20 + 10) m2 = 76 m2 (iv) Length of cuboid (l) = 3.2 m = 32 dm Breadth (b) = 30 dm Height (h) = 250 cm = 25 dm ∴ Surface area = 2(1 x b + b x h + h x l) = 2(32 x 30 + 30 x 25 + 25 x 32) dm2 = 2(960 + 750 + 800) dm2 = 2 x 2510 = 5020 dm2
Question 2. Find the surface area of a cube whose edge is (i) 1.2 m (ii) 27 cm (iii) 3 cm (iv) 6 m (v) 2.1m Solution: (i) Edge of the cube (a) = 1.2 m ∴ Surface area = 6a2= 6 x (1,2)2 m2 = 6 x 1.44 = 8.64 m2 (ii) Edge of cube (a) = 27 cm ∴ Surface area = 6a2 = 6 x (27)2 m2 = 6 x 729 = 4374 m2 (iii) Edge of cube (a) = 3 cm Surface area = 6a2 = 6 x (3)2 m2 = 6×9 cm2 = 54 cm2 (iv) Edge of cube (a) = 6 m ∴ Surface area = 6a2 = 6 x (6)2 m2 = 6 x 6 x 6 = 216 m2 (v) Edge of the cube (a) = 2.1 m ∴ Surface area = 6a2 = 6 x (2.1)2 m2 = 6 x 4.41 = 26.46 m2
Question 3. A cuboidal box is 5 cm by 5 cm by 4 cm. Find its surface area. Solution: Length of cuboid box (l) = 5 cm Breadth (b) = 5 cm and height (h) = 4 cm ∴ Surface area = 2 (l x b + b x h + h x l) = 2 (5 x 5 + 5 x 4 + 4 x 5) cm2 = 2 (25 + 20 + 20) = 2 x 65 cm2 = 130 cm2
Question 4. Find the surface area of a cube whose volume is : (i) 343 m3 (ii) 216 dm3. Solution: (i) Volume of a cube = 343 m3
Question 5. Find the volume of a cube whose surface area is (i) 96 cm2 (ii) 150 m2. Solution: (i) Surface area of a cube = 96 cm2
Question 6. The dimensions of a cuboid are in the ratio 5:3:1 and its total surface area is 414 m2. Find the dimensions. Solution: Ratio in .dimensions = 5 : 3 : 1 Let length (l) = 5x breadth (b) = 3x and height (h) = x ∴ Surface area = 2(1 x b + b x h + h x l) = 2(5x x 3x + 3x x x + x x 5x) = 2(15×2 + 3×2 + 5×2) = 2 x 23×2 = 46×2
Question 7. Find the area of the cardboard required to make a closed box of length 25 cm, 0.5 m and height 15 cm. Solution: Length of cardboard (l) = 25 cm Breadth (b) = 0.5 m = 50 cm Height (h)= 15 cm. ∴ Surface area of cardboard = 2 (l x b + b x h + h x l) = 2(25 x 50 + 50 x 15 + 15 x 25) cm2 = 2(1250+ 750+ 375) cm2 = 2(2375) = 4750 cm2
Question 8. Find the surface area of a wooden box whose shape is of a cube and if the edge of the box is 12 cm. Solution: Edge of cubic wooden box = 12 cm ∴ Surface area = 6a2 = 6(12)2 cm2 = 6 x 144 = 864 cm2
Question 9. The dimensions of an oil tin are 26 cm x 26 cm x 45 cm. Find the area of the tin sheet required for making 20 such tins. If 1 square metre of the tin sheet costs Rs 10, find the cost of the tin sheet used for these 20 tins. Solution: Length of tin (l) = 26 cm = 0.26 m Breadth (b) = 26 cm = 0.26 m Height (h) = 45 cm = 0.45 m ∴ Surface area = 2(l x b + b x h +h xl) = 2(0.26 x 0.26 + 0.26 x 0.45 + 0.45 x 0.26) m2 = 2(0.0676 + 0.117 + 0.117) m2 = 2(0.3016) = 0.6032 m2 Sheet required for such 20 tins = 0.6032 x 20= 12.064 m2 Cost of 1 m2 tin sheet = 10 m ∴ Total cost = Rs 12.064 x 10 = Rs 120.64 and area of sheet = 12.064 m2 = 120640 cm2
Question 10. A classroom is 11 m long, 8 m wide and 5 m high. Find the sum of the areas of its floor and the four walls (including doors, windows etc.) Solution: Length of room (l) = 11 m Width (b) = 8 m and height (h) = 5 m Area of floor = l x b = 11 x8 = 88m2 Area of four walls = 2 (l + b) x h = 2(11 + 8) x 5 m2 = 2 x 19×5 = 190 m2 ∴ Total area = 88 m2 + 190 m2 = 278 m2
Question 11. A swimming pool is 20 m long, 15 m wide and 3 m deep. Find the cost of repairing the floor and wall at the rate of Rs 25 per square metre. Solution: Length of pool (l) = 20 m Breadth (b) = 15 m and Depth (h) = 3 m. Area of floor = l x b = 20 x 15 = 300 m2 and area of its walls = 2(l + b) x h = 2(20 + 15) x 3 = 2 x 35 x 3 m2 = 210 m2 ∴ Total area = 300 + 210 = 510 m2 Rate of repairing it = Rs 25 per sq. me
Question 12. The perimeter of a floor of a room is 30 m and its height is 3 m. Find the area of four walls of the room. Solution: Perimeter of floor = 30 m i.e. 2(1 + b) = 30 m Height = 3 m ∴ Area of four walls = Perimeter x height = 30 x 3 = 90 m2
Question 13. Show that the product of the areas of the floor and two adjacent walls of a cuboid is the square of its volume. Solution: Let length of the room = l and breadth = b and height = h Volume = l x b x h Area of floor = l x b = lb. Area of two adjacent walls = hl x bh. ∴ Product of areas of floor and two adjacent walls of the room = lb (hi x bh) = l2b2h2 = (l.b.h)2 = (Volume)2 Hence proved
Question 14. The walls and ceiling of a room are to be plastered. The length, breadth and height of the room are 4.5, 3m and 350 cm, respectively. Find the cost of plastering at the rate of Rs 8 per square metre. Solution: Length of room (l) = 4.5 m Width (b) = 3 m and height (h) = 350 cm = 3.5 m ∴ Area of walls = 2(l + b) x h = 2(4.5 + 3) x 3.5 m2 = 2 x 7.5 x 3.5 m2 = 52.5 m2 Area of ceiling = l x b = 4.5 x 3 = 13.5 m2 ∴ Total area = 52.5 + 13.5 m2 = 66 m2 Rate of plastering = Rs 8 per sq. m ∴ Total cost = Rs 8 x 66 = Rs 528
Question 15. A cuboid has total surface area of 50 m2 and lateral surface area its 30 m2. Find the area of its base. Solution: Total surface area of cuboid = 50 m2 Lateral surface area = 30 m2 ∴ Area of floor and ceiling = 50 – 30 = 20 m2 But area of floor = area of ceiling ∴ Area of base (floor) = 202 = 10 m2
Question 16. A classroom is 7 m long, 6 m broad and 3.5 m high. Doors and windows occupy an area of 17 m2. What is the cost of white-washing the walls at the rate of Rs 1.50 per m2. Solution: Length of room (l) = 7 m Breadth (b) = 6 m and height (h) = 3.5 m ∴ Area of four walls = 2(1 + b) x h = 2(7 + 6) x 3.5 m2 = 2 x 13 x 3.5 = 91 m2 Area of doors and windows = 17 m2 ∴ Remaining area of walls = 91 – 17 = 74 m2 Rate of whitewashing = Rs 1.50 per m2 ∴ Total cost = 74 x Rs 1.50 = Rs 111
Question 17. The central hall of a school is 80 m long and 8 m high. It has 10 doors each of size 3 m x 1.5 m and 10 windows each of size 1.5 m x l m. If the cost of the white-washing the walls of the hall at the rate of Rs 1.20 per m2 is Rs 2385.60, find the breadth of the hall. Solution: Length of hall (l) = 80 m Height (h) = 8 m Size of each door = 3 m x 1.5 m ∴ Area of 10 doors = 3 x 1,5 x 10 m2 = 45 m2 A size of each windows = 1.5 m x 1 m ∴ Area of 10 windows = 1.5 m x 1 x 10= 15 m2 Total cost of whitewashing the walls = Rs 2385.60 Rate of whitewashing = Rs 1.20 per m2 ∴ Area of walls which are whitewashed
Exercise 21.4
Question 1. Find the length of the longest rod that can be placed in a room 12 m long, 9 m broad and 8 m high. Solution: Length of room (l) = 12m Breadth (b) = 9 m Height (h) = 8 m Longest rod to be kept in the room
Question 2. If V is the volume of the cuboid of dimensions a, b, c and S its the surface area then prove that Solution: ∵ a, b, c are the dimensions of a cuboid ∴ Volume (V) = abc Surface area (S) = 2(ab + bc + ca) Now
Question 3. The areas of three adjacent faces of a cuboid are .v, y and z. If the volume is V1 prove that V2 = xyz. Solution: Let length of cuboid = l Breadth = b and height = h Volume = Ibh ∴ x = lb,y = bh and z = hl Now x.y.z = lb.bh.hl = l2 b2 h2 = (Ibh)2 = V2 ∴ V2 = xyz Hence proved
Question 4. A rectangular water reservoir contains 105 m3 of water. Find the depth of the water in the reservoir if its base measures 12 m by 3.5 m. Solution: Volume of the water in reservoir = 105 m2 Length (l)= 12 m and breadth (b) = 3.5 m
Question 5. Cubes A, B, C having edges 18 cm, 24 cm and 30 cm respectively are melted and moulded into a new cube D. Find the edge of the bigger cube D. Solution: Edge of cube A = 18 cm ∴ Volume = a2 = (18)3 cm3 = 5832 cm3 Edge of cube B = 24 cm ∴ Volume = (24)3 = 13824 cm3 Edge of cube C = 30 cm ∴Volume = (30)3 = 27000 cm3 Volume of A, B, C cubes = 5832+ 138-24+ 27000 = 46656 cm3 Volume of cube D = 46656 cm3
Question 6. The breadth of a room is twice its height, one half of its length and the volume of the room is 512 cu.dm. Find its dimensions. Solution: Volume of room = 512 cu.dm Let height of the room (h) = x Then breadth (b) = 2x and length (l) = 2x x 2 = 4x. ∴ Volume = l x b x h = 4x x 2x x x = 8×3
Question 7. A closed iron tank 12 m long, 9 m wide and 4 m deep is to be made. Determine the cost of iron sheet used at the rate of Rs 5 per metre sheet, sheet being 2 m wide. Solution: Length of iron tank (l) = 12 m Breadth (b) = 9 m Depth (h) = 4 m ∴ Surface area of the tank = 2(l x b + b x h + h x l) = 2(12 x 9 + 9 x 4 + 4 x 12) m2 = 2(108 + 36 + 48) = 2 x 192 m2 = 384 m2 Width of sheet used = 2 m
Question 8. A tank open at the top is made of iron sheet 4 m wide. If the dimensions of the tank are 12 m x 8 m x 6 m, find the cost of iron sheet at Rs 17.50 per metre. Solution: Dimensions of the open iron tank = 12mx 8m.x 6m ∴ Surface area (without top) = 2(1 x b) x h + lb = 2(12 + 8) x 6+12 x 8m2 = 2 x 20 x 6 + 96 = 240 + 96 m2 = 336 m2 Width of sheet used = 4 m ∴ Length of sheet = Areab = 3364 m = 84 m b 4 Rate of sheet = Rs 17.50 per m. ∴ Total cost = Rs 17.50 x 84 = Rs 1470
Question 9. Three equal cubes are placed adjacently in a row. Find the ratio of the total surface area of the new cuboid to that of the sum of the surface areas of the three cubes. Solution: Let edge of each equal cubes = x Then, surface area of one cube = 6x2 and surface area of three cubes = 3 x 6x2 = 18x2 By placing the cubes in a row, The length of newly formed cuboid (l) = 3x Breadth (b) = x and height (h) = x ∴ Surface area of the cuboid so formed
Question 10. The dimensions of a room are 12.5 m by 9 m by 7 m. There are 2 doors and 4 windows in the room; each door measures 2.5 m by 1.2 m and each window 1.5 m by 1 m. Find the cost of painting the walls at Rs 3.50 per square metre. Solution: Dimensions of a room = 12.5 m x 9 m x 7 m ∴ Total surface area of the walls = 2(1 + b) x h = 2(12.5 + 9) x 7 m2 = 2 x 21.5 x 7 = 301 0 m2 Area of 2 doors = 2 x (2.5 x 1.2) m2 = 2 x 3.00 = 6 m2 Area of 4 windows = 4 x (1.5 x 1) m2 4 x 1.5 = 6 m2 ∴ Remaining area of the walls = 301 -(6 + 6) m2 = 301 – 12 = 289 m2 ∴ Rate of painting the walls = Rs 3.50 per m2 ∴ Total cost = Rs 3.50 x 289 = Rs 1011.50
Question 11. A field is 150 m long and 100 m wide. A plot (outside the field) 50 m long and 30 m wide is dug to a depth of 8 m and the earth taken out from the plot is spread evenly in the field. By how much the level of field is raised ? Solution: Length of the plot (l) = 50 m Width (b) = 30 m and depth (h) = 8 m ∴ Volume of the earth dug out = l x b x h = 50 x 30 x 8 = 12000 m3 Length of the field = 150 m and breadth = 100 m ∴ Height of the earth spread out on the field
Question 12. Two cubes, each of volume 512 cm3 are joined end to end, find the surface area of the resulting cuboid. Solution: Volume of each cube = 512 cm3 Now by joining the two equal cubes of side 8 cm, the length of so formed cuboid (l) = 2 x 8 = 16 cm Breadth (b) = 8 cm and height (h) = 8 cm ∴ Surface area = 2( l x b + b x h + h x l) = 2(16 X 8 + 8 X 8 + 8X16) cm2 = 2(128 + 64 + 128) cm2 = 2 x 320 = 640 cm2
Question 13. Three cubes whose edges measure 3 cm, 4 cm and 5 cm respectively are melted to form a new cube. Find the surface area of the new cube formed. Solution: Edge of first cube = 3 cm ∴ Volume = a3 = (3)3 27 cm3 Edge of second cube = 4 cm ∴Volume = a3 = (4)3 = 64 cm3 Edge of third cube = 5 cm ∴ Volume = a3 = (5)3 = 125 cm3 Volume of three cubes together = 27 + 64+ 125 = 216 cm3 ∴ Volume of the new cube = 216 cm3
Question 14. The cost of preparing the’walls of a room 12 m long at the rate of Rs 1.35 per square metre is Rs 340.20 and the cost of matting the floor at 85 paise per square metre is Rs 91.80. Find the height of the room. Solution: Length of the room (l) = 12 m Rate of matting the floor = 85 paise per m2 Total cost of matting = Rs 91.80
Question 15. The length of a hall is 18 m and width 12 m. The sum of the areas of the floor and the flat roof is equal to the sum of the areas of the four walls. Find the height of the wall. Solution: Length of hall (l) = 18 m and breadth (b) = 12 m ∴ Area of floor = l x b = 18 x 12 = 216 m2 and area of roof = 216 m2 Total area of floor and roof = (216 + 216) m2 = 432 m2 ∴ Area of four walls = 432 m2 But area of 4 walls = 2(l + b) x h ∴ 2h (l + b) = 432 ⇒ 2h (18 + 12) = 432 ⇒ 2h x 30 = 432 432 ⇒ h = 43260 = 7.2m ∴ Height of the wall = 7.2 m
Question 16. A metal cube of edge 12 cm is melted and formed into three smaller cubes. If the edges of the two smaller cubes are 6 cm and 8 cm, find the edge of the third smaller cube. Solution: Edge of metal bigger cube = 12 cm ∴ Volume = (12)3 = 1728 cm3 ∴ Sum of volumes of 3 smaller cubes = 1728 cm3 Edge of first smaller cube = 6 cm ∴ Volume = (6)3 = 216 cm3 Edge of second smaller cube = 8 cm ∴ Volume = (8)3 = 512 cm3 Sum of volumes of two smaller cubes = 216+ 512 = 728 cm3 ∴ Volume of third smaller cube = 1728-728 cm3 = 1000 cm3
Question 17. The dimensions of a cinema hall are 100 m, 50 m and 18 m. How many persons can sit in the hall if each person requires 150 m3 of air ? Solution: Length of cinema hall (l) = 100 m Breadth (b) = 50 m and height (h) = 18 m ∴ Volume of air of the hall = l x b x h = 100 x 50 x 18 m3 = 90000 m3 Each person requires air = 150 m3 ∴ Number of persons = 90000150= 600
Question 18. The external dimensions of a closed wooden box are 48 cm, 36 cm, 30 cm. The box is made of 1.5 cm thick wood. How many bricks of size 6 cm x 3 cm x 0.75 cm can be put in this box ? Solution: Outer dimensions of a closed wooden box = 48 cm x 36 cm x 30 cm Thickness of wood = 1.5 cm. ∴ Inner length (l) = 48 – 2 x 1.5 cm = 48 – 3 = 45 cm Breadth(b) = 36-2 x 1.5 = 36-3 = 33 cm Height (h) = 30 – 2 x 1.5 = 30 – 3 = 27 cm ∴ Volume of inner box = l x b x h = 45 x 33 x 27 cm3 = 40095 cm3 Volume of one brick of size 6 cm x 3 cm x 0.75 cm = 6 x 3 x 0.75 = 6 x 3 x 34 cm3 = 272 cm3 ∴ Number of bricks = 40095×227 = 1485 x 2 = 2970 bricks
Question 19. The dimensions of a rectangular box are in the ratio of 2 : 3 : 4 and the difference between the cost of covering it with sheet of paper at the rates of Rs 8 and Rs 9.50 per m2 is Rs 1,248. Find the dimensions of the box. Solution: Ratio in the dimensions of a box =2:3:4 Difference in total cost = Rs 1,248 Difference in rates = Rs 9.50 – Rs 8 = Rs 1.50 Let length (l) = 2x Then breadth (b) = 3x and height (h) = 4x ∴ Surface area = 2 (l x b + b x h + h x l) = 2(2x 3x + 3x x 4x + 4x x 2x) = 2(6x2 + 12x2 + 8 x2) = 2 x 26x2 = 52x2 First rate of paper = Rs 9.50 per m2 and second rate = 8.00 per m2 ∴ First cost = Rs 52x2 x 9.50 and second cost = Rs 52x2 x 8 ∴ 52x2 x 9.50 – 52x2 x 8= 1248 ⇒ 52x2 (9.50 – 8) = 1248 ⇒ 52x2(1.50) = 1248
Question 1. A flooring tile has the shape of a parallelogram whose base is 24 cm and the corresponding height is 10 cm. How many such tiles are required to cover a floor of area 1080 m² ? Solution: Area of floor = 1080 m² Base of parallelogram shaped tile (b) = 24 cm and corresponding height (h) = 10 cm Area of one tile = b x h = 24 x 10 = 240 cm²
Question 2. A plot is in the form of a rectangle ABCD having semi-circle on BC as shown in Fig. If AB = 60 m and BC = 28 m, find the area of the plot. Solution: Length of rectangular portion (l) = 60 m and breadth (b) = 28 m Area of the rectangular plot = l x b = 60 x 28 m² = 1680 m² Radius of semicircular portion (r) = b2 = 282 = 14 m Area = 12 πr² = 12 x 227 x 14 x 14 m² = 308 m² Total area of the plot = 1680 + 308 = 1988 m²
Question 3. A playground has the shape of a rectangle, with two semi-circles on its smaller sides as diameters, added to its outside. If the sides of the rectangle are 36 m and 24.5 m, find the area of the playground. (Take π = 227). Solution: Length of rectangular portion (l) = 36 m and breadth (b) = 24.5 m = 227 x 150.0625 m² = 471.625 m² Total area of the playground = 471.625 + 882 = 1353.625 m²
Question 4. A rectangular piece is 20 m long and 15 m wide. From its four corners, quadrants of radii 3.5 have been cut. Find the area of the remaining part. Solution: Length of rectangular piece (l) = 20 m breadth (b) = 15 m Area of rectangular piece = l x b = 20 x 15 = 300 m² Radius of each quadrant (r) = 3.5 m Total area of 4 quadrants Area of the remaining portion = 300 – 38.5 m² = 261.5 m²
Question 5. The inside perimeter of a running track (shown in Fig.) is 400 m. The length of each of the straight portion is 90 m and the ends are semi-circles. If track is everywhere 14 m wide, find the area of the track. Also, find the length of the outer running track. Solution: Inner perimeter = 400 m. Length (l) = 90 m. Perimeter of two semicircles = 400 – 2 x 90 = 400 – 180 = 220 m
Question 6. Find the area of the Figure in square cm, correct to one place of decimal. (Take π = 227) Solution: Length of square (a) = 10 cm. Area = a² = (10)² = 100 cm² Base of the right triangle AED = 8 cm and height = 6 cm
Question 7. The diameter of a wheel of a bus is 90 cm which makes 315 revolutions per minute. Determine its speed in kilometres per hour. (Take π = 227) Solution: Diameter of the wheel (d) = 90 cm.
Question 8. The area of a rhombus is 240 cm² and one of the diagonal is 16 cm. Find another diagonal. Solution: Area of rhombus = 240 cm² Length of one diagonal (d1) = 16 cm Second diagonal (d2)
Question 9. The diagonals of a rhombus are 7.5 cm and 12 cm. Find its area. Solution: In rhombus, diagonal (d1) = 7.5 cm and diagonal (d2) = 12 cm
Question 10. The diagonal of a quadrilateral shaped field is 24 m and the perpendiculars dropped on it from the remaining opposite vertices are 8 m and 13 m. Find the area of the field. Solution: In quadrilateral shaped field ABCD, diagonal AC = 24 m and perpendicular BL = 13 m and perpendicular DM on AC = 8 m Area of the field ABED = 12 x AC x (BL + DM) = 12 x 24 x (13 + 8) m² = 12 x 21 = 252 m²
Question 11. Find the area of a rhombus whose side is 6 cm and whose altitude is 4 cm. If one of its diagonals is 8 cm long, find the length of the other diagonal. Solution: Side of rhombus (b) = 6 cm Altitude (h) = 4 cm
Question 12. The floor of a building consists of 3000 tiles which are rhombus shaped and each of its diagonals are 45 cm and 30 cm in length. Find the total cost of polishing the floor, if the cost per m² is Rs 4. Solution: Number of rhombus shaped tiles = 300 Diagonals of each tile = 45 cm and 130 cm Rate of polishing the tiles = Rs 4 per m² Total cost = 202.5 x 4 = Rs 810
Question 13. A rectangular grassy plot is 112 m long and 78 broad. It has a gravel path 2.5 m wide all around it on the side. Find the area of the path and the cost of constructing it at Rs 4.50 per square metre. Solution: Length of rectangular plot (l) = 112 m and breadth (b) = 78 m Width of path = 2.5 m Inner length = 112 – 2 x 2.5 = 112 – 5 = 107 m and inner breadth = 78 – 2 x 2.5 = 78 – 5 = 73 m Area of path = outer area – inner area = (112 x 78 – 107 x 73) m² = 8736 – 7811 = 925 m² Rate of constructing = Rs 4.50 per m² Total cost = 925 x Rs 4.50 = Rs 4162.50
Question 14. Find the area of a rhombus, each side of which measures 20 cm and one of whose diagonals is 24 cm. Solution: Side of rhombus = 20 cm. One diagonal (d1) = 24 cm Diagonals of a rhombus bisect each other at right angle AB = 20 cm and OA = 12 AC = 12 x 24 cm = 12 cm In right-angled ∆AOB, AB² = AO² + BO² (Pythagoras theorem) ⇒ (20)² = (12)² + BO² ⇒ 400 = 144 + BO² ⇒ BO² = 400 – 144 = 256 = (16)² ⇒ BO = 16 cm and diagonal BD = 2 x BO = 2 x 16 = 32 cm Now area of rhombus ABCD
Question 15. The length of a side of a square field is 4 m. What will be the altitude of the rhombus if the area of the rhombus is equal to the square field and one of its diagonal is L m ? Solution: Side of square = 4 m Area of square = (a)² = 4 x 4 =16 m² Diagonals of a rhombus bisect each other at right angles. In right ∆AOB AB² = QA² + BO² (Pythagoras theorem) = (8)² + (1)² = 64 + 1 = 65 AB = √65 m. Now, length of perpendicular AL (h)
Question 16. Find the area of the field in the form of a rhombus, if the length of each side be 14 cm and the altitude be 16 cm. Solution: Length of each side of rhombus = 14 cm. Length of altitude = 16 cm Area = Base x altitude = 14 x 16 cm² = 224 cm²
Question 17. The cost of fencing a square field at 60 paise per metre is Rs 1,200. Find the cost of reaping the field at the rate of 50 paise per 100 sq. metres. Solution: Cost of fencing the square field = Rs 1,200 Rate = 60 paise per m.
Question 18. In exchange of a square plot one of whose sides is 84 m, a man wants to buy a rectangular plot 144 m long and of the same area as of the square plot. Find the width of the rectangular plot. Solution: Side of a square plot = 84 m Area = (a)² = (84)² = 84 x 84 m² = 7056 m² Area of rectangular field = 7056 m² Length (l) = 144 m
Question 19. The area of a rhombus is 84 m². If its perimeter is 40 m, then find its altitude. Solution: Area of rhombus = 84 m² Perimeter = 40 m
Question 20. A garden is in the form of a rhombus whose side is 30 metres and the corresponding altitude is 16 m. Find the cost of levelling the garden at the rate of Rs 2 per m². Solution: Side of rhombus garden (b) = 30 m. Altitude (h) = 16 m Area = Base x altitude = 30 x 16 = 480 m² Rate of levelling the garden = Rs 2 per m² Total cost = Rs 480 x 2 = Rs 960
Question 21. A field in the form of a rhombus has each side of length 64 m and altitude 16 m. What is the side of a square field which has the same area as that of a rhombus ? Solution: Length of side of rhombus (b) = 64 m and altitude (h) = 16 m Area = b x h = 64 x 16 m² = 1024 m² Now area of square = 1024 m² Side of the square = √Area = √1024 m = 32 m
Question 22. The area of a rhombus is equal to the area of a triangle whose base and the corresponding altitude are 24.8 cm and 16.5 cm respectively. If one of the diagonals of the rhombus is 22 cm, find the length of the other diagonal. Solution: Base of triangle (b) = 24.8 cm and altitude (h) = 16.5 cm Area of triangle = 12 x base x height = 12 x bh= 12 x 24.8 x 16.5 cm² = 204.6 cm² Area of rhombus = 204.6 cm² Length of one diagonal (d1 = 22 cm Second diagonal (d2)
Exercise 20.2
Question 1. Find the area, in square metres, of the trapezium whose bases and altitude are as under: (i) bases = 12 dm and 20 dm, altitude =10 dm (ii) bases = 28 cm and 3 dm, altitude = 25 cm (iii) bases = 8 m and 60 dm, altitude = 40 dm (iv) bases = 150 cm and 30 dm, altitude = 9 dm Solution:
Question 2. Find the area of trapezium with base 15 cm and height 8 cm. If the side parallel to the given base is 9 cm long. Solution: In the trapezium ABCD,
Question 3. Find the area of a trapezium whose parallel sides are of length 16 dm and 22 dm and whose height is 12 dm. Solution: Length of parallel sides of a trapezium are 16 dm and 22 dm i.e. b1 = 16 dm, b2 = 22 dm and height (h) = 12 dm
Question 4. Find the height of a trapezium, the sum of lengths of whose bases (parallel sides) is 60 cm and whose area is 600 cm² Solution: Sum of parallel sides (b1 + b2) = 60 cm Area of trapezium = 600 cm²
Question 5. Find the altitude of a trapezium whose area is 65 cm² and whose bases are 13 cm and 26 cm. Solution: Area of a trapezium = 65 cm² Bases are 13 cm and 26 cm i.e. b1 = 13 cm, b2 = 26 cm.
Question 6. Find the sum of the lengths of the bases of trapezium whose area is 4.2 m² and whose height is 280 cm. Solution: Area of trapezium = 4.2 m² Height (h) = 280 cm = 2.8 m.
Question 7. Find the area of a trapezium whose parallel sides of lengths 10 cm and 15 cm are at a distance of 6 cm from each other. Calculate the area as (i) the sum of the areas of two triangles and one rectangle. (ii) the difference of the area of a rectangle id the sum of the areas of two triangles. Solution: In trapezium ABCD, parallel sides or bases are 10 cm and 15 cm and height = 6 cm Area of trapezium
Area of trapezium = 90 – 15 = 75 cm² = area of rectangle – areas of two triangles.
Question 8. The area of a trapezium is 960 cm². If the parallel sides are 34 cm and 46 cm, find the distance between them: Solution: Area of trapezium = 960 cm² Parallel sides are 34 cm and 46 cm b1 + b2 = 34 + 46 = 80 cm Distance between parallel sides = 24 cm
Question 9. Find the area of the figure as the sum of the areas of two trapezium and a rectangle. Solution: In the figure, One rectangle is ABCD whose sides are 50 cm and 10 cm. Two trapezium of equal size in which parallel sides are 30 cm and 10 cm and height
Question 10. Top surface of a table is trapezium in shape. Find its area if its parallel sides are 1 m and 1.2 m and perpendicular distance between them is 0.8 m. Solution: Top of a table is of trapezium in shape whose parallel sides are 1 m and 1.2 m and distance between them (h) = 0.8 m Area of trapezium = \(\frac { 1 }{ 2 }\) (Sum of parallel sides) x height
Question 11. The cross-section of a canal is a trapezium in shape. If the canal is 10 m wide at the top 6 m wide at the bottom, and the area of the cross-section is 72 m², determine its depth. Solution: Area of cross-section = 72 m² Parallel sides of the trapezium = 10 m and 6 m
Question 12. The area of a trapezium is 91 cm² and its height is 7 cm. If one of the parallel sides is longer than the other by 8 cm, find the two parallel sides. Solution: Area of trapezium = 91 cm² Height (h) = 7 cm. Sum of parallel sides One parallel side = 9 cm and second side = 9 + 8 = 17 cm Hence parallel sides are 17 cm, 9 cm
Question 13. The area of a trapezium is 384 cm². Its parallel sides are in the ratio 3 : 5 and the perpendicular distance between them is 12 cm. Find the length of each one of the parallel sides. Solution: Area of trapezium = 384 cm² Perpendicular distance (h) = 12 cm Sum of parallel sides First parallel side = 8 x 3 = 24 cm Second side = 8 x 5 = 40 cm
Question 14. Mohan wants to buy a trapezium shaped field. Its side along the river is parallel and twice the side along the road. If the area of this field is 10500 m² and the perpendicular distance between the two parallel sides is 100 m, find the length of the side along the river. Solution: Area of the trapezium shaped field = 10500 m² and perpendicular distance between them (h) = 100 m.
Question 15. The area of trapezium is 1586 cm² and the distance between the parallel sides is 26 cm. If one of the parallel sides is 38 cm, find the other. Solution: Area of a trapezium = 1586 cm² and distance between the parallel sides (h) = 26
Question 16. The parallel sides of a trapezium are 25 cm and 13 cm ; Its nonparallel sides are equal each being 10 cm, find the area of the trapezium. Solution: Parallel sides of a trapezium ABCD are 25 cm and 13 cm i.e. AB = 25 cm, CD = 13 cm and each non-parallel side = 10 cm i.e., AD = BC = 10 cm From C, draw CE || DA and draw CL ⊥ AB CE = DA = CB = 10 cm and EB = AB – AE = AB – DC = 25 – 13 = 12 cm Perpendicular CL bisects base EB of an isosceles ∆CED
Question 17. Find the area of a trapezium whose parallel sides are 25 cm, 13 cm and other sides are 15 cm each. Solution: In trapezium ABCD, parallel sides are AB and DC. AB = 25 cm, CD = 13 cm and other sides are 15 cm each i.e. AD = CB = 15 cm From C, draw CE || DA and CL ⊥ AB AE = DC = 13 cm and EB = AB – AE = 25 – 13 = 12 cm Perpendicular CL bisects the base EB of the isosceles triangle CEB
Question 18. If the area of a trapezium is 28 cm² and one of its parallel sides is 6 cm, find the other parallel side if its altitude is 4 cm. Solution: Area of trapezium = 28 cm² Altitude (h) = 4 cm. One of the parallel side = 6 cm Second parallel side = 14 – 6 = 8 cm
Question 19. In the figure, a parallelogram is drawn in a trapezium the area of the parallelogram is 80 cm², find the area of the trapezium. Solution: Area of parallelogram (AECD) = 80 cm² Side AE (b) = 10 cm
Question 20. Find the area of the field shown in the figure by dividing it into a square rectangle and a trapezium. Solution: Produce EF to H to meet AB at H and draw DK || EH HF = 4 cm, KD = HE = 4 + 4 = 8 cm HK = ED = 4 cm, KB = 12 – (8) = 4 cm Now, area of square AGFH = 4 x 4 = 16 cm² area of rectangle KDEH = l x b = 8 x 4 = 32 cm² and area of trapezium BCDK. Total area of the figure = 16 + 32 + 22 = 70 cm²
Exercise 20.3
Question 1. Find the area of the pentagon shown in the figure if AD = 10 cm, AG = 8 cm, AH = 6 cm, AF = 5 cm and BF = 5 cm, CG = 7 cm and EH = 3 Solution: In the figure, here are three triangles and one trapezium.
Question 2. Find the area enclosed by each of the following figures as the sum of the areas of a rectangle and a trapezium: Solution: (i) In the figure ABCDEF, Join CF, then, the figure consists one square and one trapezium ABCF is a square whose side = 18 cm Area of the square = 18 x 18 cm² = 324 cm² Area of trapezium FCDE = 12 (CF + ED) x 8 cm² = 12 (18 + 7) x 8 = 12 x 25 x 8 cm² = 100 cm² Total area of fig. ABCDEF = 324 + 100 = 424 cm² (ii) In the figure ABCDEF, Join BE. The figure consists of one rectangle BCDE and one trapezium ABEF Area of rectangle BCDE = BC x CD = 20 x 15 = 300 cm² Area of trapezium ABEF, = 12 (BE + AF) x height = 12 (15 + 6) x 8 cm² = 12 x 21 x 8 cm² = 84 cm² Area of the figure ABCDEF = 300 + 84 = 384 cm² (iii) In the figure ABCDEFGH, HC= AB = EF = 6 cm AH = BC = 4 cm DE = GF = 5 cm Join HC. In right ∆CDE, ED² = CD² + CE² ⇒ (5)² = (4)² + (CE)² ⇒ 25 = 16 + (CE)² ⇒ (CE)² = 25 – 16 = 9 = (3)² CE = 3 cm The figure consist a rectangle and a trapezium Area of rectangle ABCH = AB x BC = 6 x 4 = 24 cm² Area of trapezium GDEF, = 12 (GD + EF) x CE 1 = 12 (GH + HC + CD + EF) x CE = 12 (4 + 6 + 4 + 6) x 3 cm² = 12 x 20 x 3 cm² = 30 cm² Total area of the figure ABCDEFGH = 24 + 30 = 54 cm²
Question 3. There is a pentagonal shaped park as shown in the figure. Jyoti and Kavita divided it in two different ways. Find the area of this park using both ways. Can you suggest some another way of finding its area ? Solution: In first case, the figure ABCDE is divided into 2 trapezium of equal area. Now area of trapezium DFBC Total area of the pentagon ABCDE = 2 x 168.75 = 337.5 m² In second case, the figure ABCDE is divided into two parts, namely one square and other triangle. Total area of pentagon ABCDE = 225 + 112.5 = 337.5 m²
Question 4. Find the area of the following polygon, if AL = 10 cm, AM = 20 cm, AN = 50 cm, AO = 60 cm and AD = 90 cm. Solution: In the figure ABCDEF, AD = 90 cm BL = 30 cm AO = 60 cm CN = 40 cm AN = 50 cm EO = 60 cm AM = 20 cm FM = 20 cm AL = 10 cm Area of ABCDEF = (150 + 800 + 900 + 200 + 1400 + 1600) cm² = 5050 cm²
Question 5. Find the area of the following regular hexagon: Solution: In the regular hexagon MNOPQR There are two triangles and one rectangle. Join MQ, MO and RP NQ = 23 cm, NA = BQ = 102 = 5 cm MR = OP = 13 cm In right ∆BDQ, PQ² = BQ² + BP² ⇒ (13)² = (5)² + BP² ⇒ 169 = 25 + BP² ⇒ BP² = 169 – 25 = 144 = (12)² BP = 12 cm PR = MO = 2 x 12 = 24 cm Now area of rectangle RPOM = RP x PO = 24 x 13 = 312 cm² Area of ∆PRQ = 12 x PR x BQ = 12 x 24 x 5 = 60 cm² Similarly area ∆MON = 60 cm² Area of the hexagon MNOPQR = 312 + 60 + 60 = 432 cm²
Question 1. Construct a quadrilateral ABCD in which AB = 4.4 cm, BC = 4 cm, CD = 6.4 cm, DA = 3.8 cm and BD = 6.6 cm. Solution: Steps of construction : (i) Draw a line segment AB = 4.4 cm. (ii) With centre A and radius 3.8 cm and with centre B and radius 6.6 cm, draw arcs intersecting each other at D. (iii) With centre B and radius 4 cm, and with centre D and radius 6.4 cm, draw arcs intersecting each other at C on the other side of BD. (iv) Join AD, BD, BC and DC. The ABCD is the required quadrilateral.
Question 2. Construct a quadrilateral ABCD such that AB = BC = 5.5 cm, CD = 4 cm, DA = 6.3 cm and AC = 9.4 cm. Measure BD. Solution: (i) Draw a line segment AC = 9.4 cm. (ii) With centre A and C and radius 5.5 cm, draw arcs intersecting each other at B. (iii) Join AB and CB. (iv) Again with centre A and radius 6.3 cm, and with centre C and radius 4 cm, draw arcs intersecting each other at D below the line segment AC. (v) Join AD and CD. Then ABCD is the required quadrilateral. On measuring BD, it is 5 cm.
Question 3. Construct a quadrilateral XYZW in which XY = 5 cm, YZ = 6 cm, ZW = 7 cm, WX = 3 cm and XZ = 9 cm. Solution: Steps of construction : (i) Draw a line segment XZ = 9 cm. (ii) With centre X and radius 3 cm and with centre Z and radius 7 cm, draw arcs intersecting each other at W. (iii) Join XW and ZW. (iv) Again with centre X and radius 5 cm and with centre Z and radius 6 cm, draw arcs, intersecting each other at Y below the line segment XZ. (v) Join XY and ZY. Then XYZW is the required quadrilateral.
Question 4. Construct a parallelogram PQRS such that PQ = 5.2 cm, PR = 6.8 cm and QS = 8.2 cm. Solution: Steps of construction: In a parallelogram, diagonals bisect each other. Now (i) Draw a line segment PQ = 5.2 cm. (ii) With centre P and radius 3.4 cm (12 of PR) and with centre Q and radius 4.1 cm (12 of QS) draw arcs intersecting each other at O. (iii) Join PQ and QO and produced them to R and S respectively such that PO = OR and QO = OS. (iv) Join PS, SR and RQ. Then PQRS is the required parallelogram.
Question 5. Construct a rhombus with side 6 cm and one diagonal 8 cm. Measure the other diagonal. Solution: Steps of construction : Sides of a rhombus are equal. (i) Draw a line segment AC = 8 cm. (ii) With centres A and C and radius 6 cm, draw two arcs above the line segment AC and two below the line segment AC, intersecting each other at D and B respectively. (iii) Join AB, AD, BC and CD. Then ABCD is the required rhombus. JoinBD. On measuring BD, it is approximately 9 cm
Question 6. Construct a kite ABCD in which AB = 4 cm, BC = 4.9 cm and AC = 7.2 cm. Solution: Steps of construction : (i) Draw a line segment AC = 7.2 cm. (ii) With centre A and radius 4 cm draw an arc. (iii) With centre C and radius 4.9 cm, draw another arc which intersects the first arc at B and D. (iv) Join AB, BC, CD and DA. Then ABCD is the required kite.
Question 7. Construct, if possible, a quadrilateral ABCD given, AB = 6 cm BC = 3.7 cm, CD = 5.7 cm, AD = 5.5 cm and BD = 6.1 cm. Give reasons for not being able to construct, if you cannot. Solution: Steps of construction : (i) Draw a line segment BD = 6.1 cm. (ii) With centre B and radius 6 cm and with centre D and radius 5.5 cm, draw arcs intersecting at A. (iii) Join AB and AD. (iv) Again with centre B and radius 3.7 cm and with centre D and radius 5.7 cm, draw two arcs intersecting each other at C below the BD. (v) Join BC and DC. Then ABCD is the required quadrilateral.
Question 8. Construct, if possible a quadrilateral ABCD in which AB = 6 cm, BC = 7 cm, CD = 3 cm, AD = 5.5. cm and AC = 11 cm. Give reasons for not being able to construct, if you cannot. Solution: Steps of construction: It is not possible to construct this quadrilateral ABCD because AD + DC = 5.5 cm + 3 cm = 8.5 cm and AC = 11 cm AD + DC < AC. But we know that in a triangle, Sum of two sides is always greater than its third side.
Exercise 18.2
Question 1. Construct a quadrilateral ABCD in which AB = 3.8 cm, BC = 3.0 cm, AD = 2.3 cm, AC = 4.5 cm and BD = 3.8 cm. Solution: Steps of construction : (i) Draw a line segment AB = 3.8 cm. (ii) With centre A and radius 2.3 cm and with centre B and radius 3.8 cm draw arcs intersecting each other at D. (iii) Join AD and BD. (iv) Again with centre A and radius 4.5 cm and with centre B and radius 3 cm, draw arcs intersecting each other at C. (v) Join AC and BC and also CD. Then ABCD is the required quadrilateral.
Question 2. Construct a quadrilateral ABCD in which BC = 7.5 cm, AC = AD = 6 cm, CD = 5 cm and BD = 10 cm. Solution: Steps of construction : (i) Draw a line segment CD = 5 cm. (ii) With centre C and D and radius 6 cm, draw line segments intersecting each other at A. (iii) Join AC and AD. (iv) Again with centre C and radius 7.5 cm and with centre D and radius 10 cm, draw arcs intersecting each other at B. (v) Join CB, CA, DA, DB and AB. Then ABCD is the required quadrilateral.
Question 3. Construct a quadrilateral ABCD, when AB = 3 cm, CD = 3 cm, DA = 7.5 cm, AC = 8 cm and BD = 4 cm. Solution: Steps of construction : This quadrilateral is not possible as BD = 4 cm, AB = 3 cm and AD = 7.5 cm The sum of any two sides of a triangle is greater than the third side. But BD + AD = 4 + 3 = 7 cm BD + AD < AD
Question 4. Construct a quadrilateral ABCD given AD = 3.5 cm, BC = 2.5 cm, CD = 4.1 cm, AC = 7.3 cm and BD = 3.2 cm. Solution: Steps of construction : (i) Draw a line segment CD = 4.1 cm. (ii) With centre C and radius 7.3 cm and with centre D and radius 3.5 cm, draw arcs intersecting each other at A. (iii) Join AC and AD. (iv) Again with centre C and radius 2.5 cm and with centre D and radius 3.2 cm, draw arcs intersecting each other at B. (v) Join CB’, and DB’ and join AB’. Then ABCD is the required quadrilateral.
Question 5. Construct a quadrilateral ABCD given AD = 5 cm, AB = 5.5 cm, BC = 2.5 cm, AC = 7.1 cm and BD = 8 cm. Solution: Steps of construction: (i) Draw a line segment AB = 5 cm. (ii) With centre A and radius 7.1 cm and with centre B and radius 2.5 cm, draw arcs which intersect each other at C. (iii) Join AC and BC. (iv) Again with centre A and radius 5 cm and with centre B and radius 8 cm, draw arcs which intersect each other at D. (v) Join AD and BD and CD. Then ABCD is the required quadrilateral.
Question 6. Construct a quadrilateral ABCD in which BC = 4 cm, CA = 5.6 cm, AD = 4.5 cm, CD = 5 cm and BD = 6.5 cm. Solution: Steps of construction: (i) Draw a line segment CD = 5 cm. (ii) With centre C and radius 5.6 cm and with centre D and radius 4.5 cm, draw arcs which intersect each other at A. (iii) Join AC and AD. (iv) Again with centre C and radius 4 cm and with centre D and radius 6.5 cm, draw arcs which intersect each other at B. (v) Join BC and BD and AB. Then ABCD is the required quadrilateral.
Exercise 18.3
Question 1. Construct a quadrilateral ABCD in which AB = 3.8 cm, BC = 3.4 cm, CD = 4.5 cm, AD = 5 cm and ∠B = 80°. Solution: Steps of construction : (i) Draw a line segment AB = 3.8 cm. (ii) At B, draw a ray BX making an angle of 80° and cut off BC = 3.4 cm. (iii) With centre A and radius 5 cm and with centre C and radius 4.5 cm, draw arcs which intersect each other at D. (iv) Join CD and AD. ABCD is the required quadrilateral.
Question 2. Construct a quadrilateral ABCD given that AB = 8 cm, BC = 8 cm, CD = 10 cm, AD = 10 cm and ∠A = 45°. Solution: Steps of construction : (i) Draw a line segment AB = 8 cm. (ii) At A, draw a ray AX making an angle of 45° and cut off BC = 8 cm. (iii) With centre A and C and radius 10 cm, draw arcs intersecting each other at D. (iv) Join AD, CD. Then ABCD is the required quadrilateral.
Question 3. Construct a quadrilateral ABCD in which AB = 7.7 cm, BC = 6.8 cm, CD = 5.1 cm, AD = 3.6 cm and ∠C = 120°. Solution: Steps of construction : (i) Draw a line segment BC = 6.8 cm. (ii) At C, draw a ray CX making an angle of 120° and cut off CD = 5.1 cm. (iii) With centre B and radius 7.7 cm and with centre D and radius 3.6 cm draw arcs which intersect each other at A. (iv) Join AD and AB. Then ABCD is the required quadrilateral.
Question 4. Construct a quadrilateral ABCD in which AB = BC = 3 cm, AD = CD = 5 cm and ∠B = 120° Solution: Steps of construction : (i) Draw a line segment AB = 3 cm. (ii) At B, draw a ray BX making an angle of 120° and cut off BC = 3 cm. (iii) With centres A and C, and radius 5 cm, draw arcs intersecting each other at D. (iv) Join AD and CD. Then ABCD is the required quadrilateral.
Question 5. Construct a quadrilateral ABCD in which AB = 2.8 cm, BC = 3.1 cm, CD = 2.6 cm and DA = 3.3 cm and ∠A = 60°. Solution: Steps of construction : (i) Draw a line segment AB = 2.8 cm. (ii) At A draw a ray AX making an angle of 60° and cut off AD = 3.3 cm. (iii) With centre B and radius 3.1 cm and with centre D and radius 2.6 cm, draw arc which intersect each other at C. (iv) Join CB and CD. Then ABCD is the required quadrilateral.
Question 6. Construct a quadrilateral ABCD in which AB = BC = 6 cm, AD = DC = 4.5 cm and ∠B = 120°. Solution: Steps of construction: The construction is not possible to draw as arcs of radius 4.5 cm from A and C, do not intersect at any point.
Exercise 18.4
Question 1. Construct a quadrilateral ABCD, in which AB = 6 cm, BC = 4 cm, CD = 4 cm, ZB = 95° and ∠C = 90°. Solution: Steps of construction : (i) Draw a line segment BC = 4 cm. (ii) At B, draw a ray BX making an angle of 95° and cut off BA = 6 cm. (iii) At C, draw a ray CY making an angle of 90° and cut off CD = 4 cm. (iv) Join AD. Then ABCD is the required quadrilateral.
Question 2. Construct a quadrilateral ABCD, where AB = 4.2 cm, BC = 3.6 cm, CD = 4.8 cm, ∠B = 30° and ∠C = 150°. Solution: Steps of construction : (i) Draw a line segment BC = 3.6 cm. (ii) At B, draw a ray BX making an angle of 30° and cut of BA = 4.2 cm. (iii) At C, draw another ray CY making an angle of 150° and cut off CD = 4.8 cm. (iv) Join AD. Then ABCD is the required quadrilateral.
Question 3. Construct a quadrilateral PQRS, in which PQ = 3.5 cm, QR = 2.5 cm, RS = 4.1 cm, ∠Q = 75° and ∠R = 120°. Solution: Steps of construction : (i) Draw a line segment QR = 2.5 cm. (ii) At Q, draw a ray QX making an angle of 75° and cut off QP = 3.5 cm. (iii) At R, draw another ray RY making an angle of 120° and cut off RS = 4.1 cm. (iv) Join PS. Then PQRS is the required quadrilateral.
Question 4. Construct a quadrilateral ABCD given BC = 6.6 cm, CD = 4.4 cm, AD = 5.6 cm and ∠D = 100° and ∠C = 95°. Solution: Steps of construction : (i) Draw a line segment CD = 4.4 cm. (ii) At C, draw a ray CX making an angle of 95° and cut off CB = 6.6 cm (iii) At D, draw another ray DY making an angle of 100° and cut off DA = 5.6 cm. (iv) Join AB. Then ABCD is the required quadrilateral.
Question 5. Construct a quadrilateral ABCD in which AD = 3.5 cm, AB = 4.4 cm, BC = 4.7 cm, ∠A = 125° and ∠B = 120°. Solution: Steps of construction : (i) Draw a line segment AB 4.4 cm. (ii) At A, draw a ray AX making an angle of 125° and cut off AD = 3.5 cm. (iii) At B, draw another ray BY making an angle of 120° and cut off BC = 4.7 cm. (iv) Join CD. Then ABCD is the required quadrilateral.
Question 6. Construct a quadrilateral PQRS in which ∠Q = 45°, ∠R = 90°, QR = 5 cm, PQ = 9 cm and RS = 7 cm. Solution: Steps of construction : This quadrilateral is not possible to construct as shown in the figure.
Question 7. Construct a quadrilateral ABCD in which AB = BC = 3 cm, AD = 5 cm, ∠A = 90° and ∠B = 105°. Solution: Steps of construction : (i) Draw a line segment AB = 3 cm. (ii) At A, draw a ray AX making an angle of 90° and cut off AD = 5 cm. (iii) At B, draw another ray BY making an angle of 105° and cut off BC = 3 cm. (iv) Join CD. Then ABCD is the required quadrilateral.
Question 8. Construct a quadrilateral BDEF where DE = 4.5 cm, EF = 3.5 cm, FB = 6.5 cm and ∠F = 50° and ∠E = 100° Solution: Steps of construction : (i) Draw a line segment EF = 3.5 cm. (ii) At E, draw a ray EX making an angle of 100° and cut off ED = 4.5 cm. (iii) At F, draw another ray FY making an angle of 45° and cut off FB = 6.5 cm. (iv) Join DB. Then BDEF is the required quadrilateral
Exercise 18.5
Question 1. Construct a quadrilateral ABCD given that AB = 4 cm, BC = 3 cm, ∠A = 75°, ∠B = 80° and ∠C = 120°. Solution: Steps of construction : (i) Draw a line segment AB = 4 cm. (ii) At A draw a ray AX making an angle of 75°. (iii) At B draw another ray BY making an angle of 80° and cut off BC = 3 cm. (iv) At C, draw another ray CZ making an angle of 120° which intersects AX at D. Then ABCD is the required quadrilateral.
Question 2. Construct a quadrilateral ABCD where AB = 5.5 cm, BC = 3.7 cm, ∠A = 60°, ∠B = 105° and ∠D = 90°. Solution: ∠A = 60°, ∠B = 105° and ∠D = 90° But ∠A + ∠B + ∠C + ∠D = 360° (Sum of angles of a quadrilateral) ⇒ 60° + 105° + ∠C + 90° = 360° ⇒ 255° + ∠C = 360° ⇒ ∠C = 360° – 255° = 105° Steps of construction : (i) Draw a line segment AB = 5.5 cm. (ii) At A, draw a ray AX making an angle of (iii) At B, draw another ray BY making an angle of 105° and cut off BC = 3.7 cm. (iv) At C, draw a ray CZ making an angle of 105° which intersects AX at D. Then ABCD is the required quadrilateral.
Question 3. Construct a quadrilateral PQRS where PQ = 3.5 cm, QR = 6.5 cm, ∠P = ∠R = 105° and ∠S = 75°. Solution: ∠P = 105°, ∠R = 105° and ∠S = 75° But ∠P + ∠Q + ∠R + ∠S = 360° (Sum of angles of a quadrilateral) ⇒ 105° + ∠Q + 105° + 75° = 360° ⇒ 285° + ∠Q = 360° ⇒ ∠Q = 360° – 285° = 75° Steps of construction : (i) Draw a line segment PQ = 3.5 cm. (ii) At P, draw a ray PX making an angle of 105°. (iii) At Q, draw another ray QY, making an angle of 75° and cut off QR = 6.5 cm. (iv) At R, draw a ray RZ making an angle of 105° which intersects PX at S. Then PQRS is the required quadrilateral.
Question 4. Construct a quadrilateral ABCD when BC = 5.5 cm, CD = 4.1 cm, ∠A = 70°, ∠B = 110° and ∠D = 85°. Solution: ∠A = 70°, ∠B = 110°, ∠D = 85° But ∠A + ∠B + ∠C + ∠D = 360° (Sum of angles of a quadrilateral) ⇒ 70° + 110° + ∠C + 85° = 360° ⇒ 265° + ∠C = 360° ⇒ ∠C = 360° – 265° = 95° Steps of construction: (i) Draw a line segment BC = 5.5 cm. (ii) At B, draw a ray BX making an angle of 110°. (iii) At C, draw another ray CY making an angle of 95° and cut off CD = 4.1 cm. (iv) At D, draw a ray DZ making an angle of 85° which intersects BX at A. Then ABCD is the required quadrilateral.
Question 5. Construct a quadrilateral ABCD, where ∠A = 65°, ∠B = 105°, ∠C = 75°, BC = 5.7 cm and CD = 6.8 cm. Solution: ∠A = 65°, ∠B = 105°, ∠C = 75° But ∠A + ∠B + ∠C + ∠D = 360° (Sum of angles of a quadrilateral) ⇒ 65° + 105° + 75° + ∠D = 360° ⇒ 245° + ∠D = 360° ⇒ ∠D = 360° – 245° = 115° Steps of construction: (i) Draw a line segment BC = 5.7 cm. (ii) At B, draw a ray BX making an angle of (iii) At C draw a another ray CY making an angle of 75° and cut off CD = 6.8 cm. (iv) At D, draw a ray DZ making an angle of 115° which intersects BX at A. Then ABCD is the required quadrilateral.
Question 6. Construct a quadrilateral PQRS in which PQ = 4 cm, QR = 5 cm, ∠P = 50°, ∠Q = 110° and ∠R = 70°. Solution: Steps of construction : (i) Draw a line segment PQ = 4 cm. (ii) At P, draw a ray PX making an angle of 50°. (iii) At Q, draw another ray QY making an angle of 110° and cut off QR = 5 cm. (iv) At R, draw a ray RZ making an angle of 70° which intersects PX at S. Then PQRS is the required quadrilateral.
to get 12?
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hours daily. She devotes
hours of her time for science and mathematics. How much time does she devote for other subjects?
hours daily
If it is cut into two pieces in such a way that the length of one piece is
what is the length of the other piece?
long and
wide. Find its perimeter?
and that of science book is
Which costs more and by how much?
